bioinformatics.pdf

Introduction to
Bioinformatics
Esa Pitkänen
esa.pitkanen@cs.helsinki.fi
Autumn 2008, I period
www.cs.helsinki.fi/mbi/courses/08-09/itb
582606 Introduction to Bioinformatics, Autumn 2008

Introduction to
Bioinformatics
Lecture 1:
Administrative issues
MBI Programme, Bioinformatics courses
What is bioinformatics?
Molecular biology primer

3
How to enrol for the course?
p Use the registration system of the Computer
Science department: https://ilmo.cs.helsinki.fi
n You need your user account at the IT department (“cc
account”)
p If you cannot register yet, don’t worry: attend the
lectures and exercises; just register when you are
able to do so

4
Teachers
p Esa Pitkänen, Department of Computer Science,
University of Helsinki
p Elja Arjas, Department of Mathematics and
Statistics, University of Helsinki
p Sami Kaski, Department of Information and
Computer Science, Helsinki University of
Technology
p Lauri Eronen, Department of Computer Science,
University of Helsinki (exercises)

5
Lectures and exercises
p Lectures: Tuesday and Friday 14.15-16.00
Exactum C221
p Exercises: Tuesday 16.15-18.00 Exactum
C221
n First exercise session on Tue 9 September

6
Status & Prerequisites
p Advanced level course at the Department
of Computer Science, U. Helsinki
p 4 credits
p Prerequisites:
n Basic mathematics skills (probability calculus,
basic statistics)
n Familiarity with computers
n Basic programming skills recommended
n No biology background required

7
Course contents
p What is bioinformatics?
p Molecular biology primer
p Biological words
p Sequence assembly
p Sequence alignment
p Fast sequence alignment using FASTA and BLAST
p Genome rearrangements
p Motif finding (tentative)
p Phylogenetic trees
p Gene expression analysis

8
How to pass the course?
p Recommended method:
n Attend the lectures (not obligatory though)
n Do the exercises
n Take the course exam
p Or:
n Take a separate exam

9
p Exercises give you max. 12 points
n 0% completed assignments gives you 0 points,
80% gives 12 points, the rest by linear
interpolation
n “A completed assignment” means that
p You are willing to present your solution in the
exercise session and
p You return notes by e-mail to Lauri Eronen (see
course web page for contact info) describing the main
phases you took to solve the assignment
n Return notes at latest on Tuesdays 16.15
p Course exam gives you max. 48 points

10
p Grading: on the scale 0-5
n To get the lowest passing grade 1, you need to get at
least 30 points out of 60 maximum
p Course exam: Wed 15 October 16.00-19.00
Exactum A111
p See course web page for separate exams
p Note: if you take the first separate exam, the
best of the following options will be considered:
n Exam gives you 48 points, exercises 12 points
n Exam gives you 60 points
p In second and subsequent separate exams, only
the 60 point option is in use

11
Literature
p Deonier, Tavaré,
Waterman: Computational
Genome Analysis, an
Introduction. Springer,
2005
p Jones, Pevzner: An
Introduction to
Bioinformatics Algorithms.
MIT Press, 2004
p Slides for some lectures
will be available on the
course web page

12
Additional literature
p Gusfield: Algorithms on
strings, trees and
sequences
p Griffiths et al: Introduction
to genetic analysis
p Alberts et al.: Molecular
biology of the cell
p Lodish et al.: Molecular cell
biology
p Check the course web site

13
Questions about administrative &
practical stuff?

14
Master's Degree Programme in
Bioinformatics (MBI)
p Two-year MSc programme
p Admission for 2009-2010 in January 2009
n You need to have your Bachelor’s degree ready by
August 2009
www.cs.helsinki.fi/mbi

15
MBI programme organizers
Department of Computer Science,
Department of Mathematics and Statistics
Faculty of Science, Kumpula Campus, HY
Laboratory of Computer and
Information Science, Laboratory of
CS and Engineering,TKK
Faculty of Medicine, Meilahti Campus, HY
Faculty of Biosciences
Faculty of Agriculture and Forestry
Viikki Campus, HY

16
TKK, Otaniemi
HY, Meilahti
HY, Kumpula
HY, Viikki
Four MBI campuses

17
MBI highlights
p You can take courses from both HY and
TKK
p Two biology courses tailored specifically
for MBI
p Bioinformatics is a new exciting field, with
a high demand for experts in job market
p Go to www.cs.helsinki.fi/mbi/careers to
find out what a bioinformatician could do
for living

18
Admission
p Admission requirements
n Bachelor’s degree in a suitable field (e.g., computer
science, mathematics, statistics, biology or medicine)
n At least 60 ECTS credits in total in computer science,
mathematics and statistics
n Proficiency in English (standardized language test:
TOEFL, IELTS)
p Admission period opens in late Autumn 2009 and
closes in 2 February 2009
p Details on admission will be posted in
www.cs.helsinki.fi/mbi during this autumn

19
Bioinformatics courses in Helsinki
region: 1st period
p Computational genomics (4-7 credits, TKK)
p Seminar: Neuroinformatics (3 credits, Kumpula)
p Seminar: Machine Learning in Bioinformatics (3
credits, Kumpula)
p Signal processing in neuroinformatics (5 credits,
TKK)

20
A good biology course for computer
scientists and mathematicians?
p Biology for methodological scientists (8 credits, Meilahti)
n Course organized by the Faculties of Bioscience and Medicine
for the MBI programme
n Introduction to basic concepts of microarrays, medical genetics
and developmental biology
n Study group + book exam in I period (2 cr)
n Three lectured modules, 2 cr each
n Each module has an individual registration so you can
participate even if you missed the first module
n www.cs.helsinki.fi/mbi/courses/08-09/bfms/

21
region: 2nd period
p Bayesian paradigm in genetic bioinformatics (6
credits, Kumpula)
p Biological Sequence Analysis (6 credits, Kumpula)
p Modeling of biological networks (5-7 credits, TKK)
p Statistical methods in genetics (6-8 credits,
Kumpula)

22
region: 3rd period
p Evolution and the theory of games (5 credits, Kumpula)
p Genome-wide association mapping (6-8 credits, Kumpula)
p High-Throughput Bioinformatics (5-7 credits, TKK)
p Image Analysis in Neuroinformatics (5 credits, TKK)
p Practical Course in Biodatabases (4-5 credits, Kumpula)
p Seminar: Computational systems biology (3 credits,
Kumpula)
p Spatial models in ecology and evolution (8 credits,
Kumpula)
p Special course in bioinformatics I (3-7 credits, TKK)

23
Bioinformatics courses in Helsinki region:
4th period
p Metabolic Modeling (4 credits, Kumpula)
p Phylogenetic data analyses (6-8 credits,
Kumpula)

25
p Bioinformatics, n. The science of information
and information flow in biological systems,
esp. of the use of computational methods in
genetics and genomics. (Oxford English
Dictionary)
p "The mathematical, statistical and computing
methods that aim to solve biological problems
using DNA and amino acid sequences and
related information." -- Fredj Tekaia

26
p "I do not think all biological computing is
bioinformatics, e.g. mathematical modelling is
not bioinformatics, even when connected with
biology-related problems. In my opinion,
bioinformatics has to do with management and
the subsequent use of biological information,
particular genetic information."
-- Richard Durbin

27
What is not bioinformatics?
p Biologically-inspired computation, e.g., genetic algorithms
and neural networks
p However, application of neural networks to solve some
biological problem, could be called bioinformatics
p What about DNA computing?
http://www.wisdom.weizmann.ac.il/~lbn/new_pages/Visual_Presentation.html

28
Computational biology
p Application of computing to biology (broad
definition)
p Often used interchangeably with bioinformatics
p Or: Biology that is done with computational
means

29
Biometry & biophysics
p Biometry: the statistical analysis of biological
data
n Sometimes also the field of identification of individuals
using biological traits (a more recent definition)
p Biophysics: "an interdisciplinary field which
applies techniques from the physical sciences
to understanding biological structure and
function" -- British Biophysical Society

30
Mathematical biology
p Mathematical biology
“tackles biological
problems, but the methods
it uses to tackle them need
not be numerical and need
not be implemented in
software or hardware.”
-- Damian Counsell
Alan Turing

31
Turing on biological complexity
p “It must be admitted that the biological examples which
it has been possible to give in the present paper are very
limited.
This can be ascribed quite simply to the fact that
biological phenomena are usually very complicated.
Taking this in combination with the relatively elementary
mathematics used in this paper one could hardly expect to
find that many observed biological phenomena would be
covered.
It is thought, however, that the imaginary biological
systems which have been treated, and the principles which
have been discussed, should be of some help in
interpreting real biological forms.”
– Alan Turing, The Chemical Basis of Morphogenesis, 1952

32
Related concepts
p Systems biology
n “Biology of networks”
n Integrating different levels
of information to
understand how biological
systems work
p Computational systems biology
Overview of metabolic pathways in
KEGG database, www.genome.jp/kegg/

33
Why is bioinformatics important?
p New measurement techniques produce
huge quantities of biological data
n Advanced data analysis methods are needed to
make sense of the data
n Typical data sources produce noisy data with a
lot of missing values
p Paradigm shift in biology to utilise
bioinformatics in research

34
Bioinformatician’s skill set
p Statistics, data analysis methods
n Lots of data
n High noise levels, missing values
n #attributes >> #data points
p Programming languages
n Scripting languages: Python, Perl, Ruby, …
n Extensive use of text file formats: need
parsers
n Integration of both data and tools
p Data structures, databases

35
p Modelling
n Discrete vs continuous domains
n -> Systems biology
p Scientific computation packages
n R, Matlab/Octave, …
p Communication skills!

36
Communication skills: case 1
Biologist presents a problem
to computer scientists /
mathematicians
?
”I am interested in finding what affects the
regulation gene x during condition y and how
that relates to the organism’s phenotype.”
”Define input and output of the problem.”

37
Bioinformatician is a part
of a group that consists
mostly of biologists.

38
...biologist/bioinformatician ratio is important!

39
A group of
bioinformaticians
offers their services to
more than one group

40
p How much biology you should know?

41
Computer Science
• Programming
• Databases
• Algorithmics
Mathematics and
statistics
• Calculus
• Probability calculus
• Linear algebra
Biology & Medicine
• Basics in molecular and
cell biology
• Measurement techniques
Bioinformatics
• Biological sequence analysis
• Biological databases
• Analysis of gene expression
• Modeling protein structure and
function
• Gene, protein and metabolic
networks
• …
Prof. Juho Rousu, 2006
Where would you be in this triangle?

42
A problem involving bioinformatics?
- ”I found a fruit fly that is immune to all diseases!”
- ”It was one of these”
Pertti Jarla, http://www.hs.fi/fingerpori/

43
Molecular Biology Primer by Angela Brooks, Raymond Brown,
Calvin Chen, Mike Daly, Hoa Dinh, Erinn Hama, Robert Hinman,
Julio Ng, Michael Sneddon, Hoa Troung, Jerry Wang, Che Fung
Yung
Edited for Introduction to Bioinformatics (Autumn 2007, Summer
2008, Autumn 2008) by Esa Pitkänen

44
p Part 1: What is life made of?
p Part 2: Where does the variation in
genomes come from?

45
Life begins with Cell
p A cell is a smallest structural unit of an
organism that is capable of independent
functioning
p All cells have some common features

46
Cells
p Fundamental working units of every living system.
p Every organism is composed of one of two radically different types of
cells:
n prokaryotic cells or
n eukaryotic cells.
p Prokaryotes and Eukaryotes are descended from the same
primitive cell.
n All prokaryotic and eukaryotic cells are the result of a total of 3.5
billion years of evolution.

47
Two types of cells: Prokaryotes and
Eukaryotes

48
Prokaryotes and Eukaryotes
p According to the most
recent evidence, there
are three main
branches to the tree of
life
p Prokaryotes include
Archaea (“ancient
ones”) and bacteria
p Eukaryotes are
kingdom Eukarya and
includes plants,
animals, fungi and
certain algae Lecture: Phylogenetic trees

49
All Cells have common Cycles
p Born, eat, replicate, and die

50
Common features of organisms
p Chemical energy is stored in ATP
p Genetic information is encoded by DNA
p Information is transcribed into RNA
p There is a common triplet genetic code
p Translation into proteins involves ribosomes
p Shared metabolic pathways
p Similar proteins among diverse groups of
organisms

51
All Life depends on 3 critical molecules
p DNAs (Deoxyribonucleic acid)
n Hold information on how cell works
p RNAs (Ribonucleic acid)
n Act to transfer short pieces of information to different
parts of cell
n Provide templates to synthesize into protein
p Proteins
n Form enzymes that send signals to other cells and
regulate gene activity
n Form body’s major components (e.g. hair, skin, etc.)
n “Workhorses” of the cell

52
DNA: The Code of Life
p The structure and the four genomic letters code for all living
organisms
p Adenine, Guanine, Thymine, and Cytosine which pair A-T and C-G
on complimentary strands.
Lecture: Genome sequencing
and assembly

53
Discovery of the structure of DNA
p 1952-1953 James D. Watson and Francis H. C. Crick
deduced the double helical structure of DNA from X-ray
diffraction images by Rosalind Franklin and data on amounts
of nucleotides in DNA
James Watson and
Francis Crick
Rosalind
Franklin
”Photo 51”

54
DNA, continued
p DNA has a double helix
structure which is
composed of
n sugar molecule
n phosphate group
n and a base (A,C,G,T)
p By convention, we read
DNA strings in direction of
transcription: from 5’ end
to 3’ end
5’ ATTTAGGCC 3’
3’ TAAATCCGG 5’

55
DNA is contained in chromosomes
http://en.wikipedia.org/wiki/Image:Chromatin_Structures.png
p In eukaryotes, DNA is packed into chromatids
n In metaphase, the “X” structure consists of two identical
chromatids
p In prokaryotes, DNA is usually contained in a single,
circular chromosome

56
Human chromosomes
p Somatic cells in humans
have 2 pairs of 22
chromosomes + XX
(female) or XY (male) =
total of 46 chromosomes
p Germline cells have 22
chromosomes + either X or
Y = total of 23
chromosomes
Karyogram of human male using Giemsa staining
(http://en.wikipedia.org/wiki/Karyotype)

57
Length of DNA and number of chromosomes
Organism #base pairs #chromosomes (germline)
Prokayotic
Escherichia coli (bacterium) 4x106 1
Eukaryotic
Saccharomyces cerevisia (yeast) 1.35x107 17
Drosophila melanogaster (insect) 1.65x108 4
Homo sapiens (human) 2.9x109 23
Zea mays (corn / maize) 5.0x109 10

58
1 atgagccaag ttccgaacaa ggattcgcgg ggaggataga tcagcgcccg agaggggtga
61 gtcggtaaag agcattggaa cgtcggagat acaactccca agaaggaaaa aagagaaagc
121 aagaagcgga tgaatttccc cataacgcca gtgaaactct aggaagggga aagagggaag
181 gtggaagaga aggaggcggg cctcccgatc cgaggggccc ggcggccaag tttggaggac
241 actccggccc gaagggttga gagtacccca gagggaggaa gccacacgga gtagaacaga
301 gaaatcacct ccagaggacc ccttcagcga acagagagcg catcgcgaga gggagtagac
361 catagcgata ggaggggatg ctaggagttg ggggagaccg aagcgaggag gaaagcaaag
421 agagcagcgg ggctagcagg tgggtgttcc gccccccgag aggggacgag tgaggcttat
481 cccggggaac tcgacttatc gtccccacat agcagactcc cggaccccct ttcaaagtga
541 ccgagggggg tgactttgaa cattggggac cagtggagcc atgggatgct cctcccgatt
601 ccgcccaagc tccttccccc caagggtcgc ccaggaatgg cgggacccca ctctgcaggg
661 tccgcgttcc atcctttctt acctgatggc cggcatggtc ccagcctcct cgctggcgcc
721 ggctgggcaa cattccgagg ggaccgtccc ctcggtaatg gcgaatggga cccacaaatc
781 tctctagctt cccagagaga agcgagagaa aagtggctct cccttagcca tccgagtgga
841 cgtgcgtcct ccttcggatg cccaggtcgg accgcgagga ggtggagatg ccatgccgac
901 ccgaagagga aagaaggacg cgagacgcaa acctgcgagt ggaaacccgc tttattcact
961 ggggtcgaca actctgggga gaggagggag ggtcggctgg gaagagtata tcctatggga
1021 atccctggct tccccttatg tccagtccct ccccggtccg agtaaagggg gactccggga
1081 ctccttgcat gctggggacg aagccgcccc cgggcgctcc cctcgttcca ccttcgaggg
1141 ggttcacacc cccaacctgc gggccggcta ttcttctttc ccttctctcg tcttcctcgg
1201 tcaacctcct aagttcctct tcctcctcct tgctgaggtt ctttcccccc gccgatagct
1261 gctttctctt gttctcgagg gccttccttc gtcggtgatc ctgcctctcc ttgtcggtga
1321 atcctcccct ggaaggcctc ttcctaggtc cggagtctac ttccatctgg tccgttcggg
1381 ccctcttcgc cgggggagcc ccctctccat ccttatcttt ctttccgaga attcctttga
1441 tgtttcccag ccagggatgt tcatcctcaa gtttcttgat tttcttctta accttccgga
1501 ggtctctctc gagttcctct aacttctttc ttccgctcac ccactgctcg agaacctctt
1561 ctctcccccc gcggtttttc cttccttcgg gccggctcat cttcgactag aggcgacggt
1621 cctcagtact cttactcttt tctgtaaaga ggagactgct ggccctgtcg cccaagttcg
1681 ag
Hepatitis delta virus, complete genome

59
RNA
p RNA is similar to DNA chemically. It is usually only a
single strand. T(hyamine) is replaced by U(racil)
p Several types of RNA exist for different functions in
the cell.
http://www.cgl.ucsf.edu/home/glasfeld/tutorial/trna/trna.gif
tRNA linear and 3D view:

60
DNA, RNA, and the Flow of
Information
Translation
Transcription
Replication
”The central dogma”
Is this true?
Denis Noble: The principles of Systems Biology illustrated using the virtual heart
http://velblod.videolectures.net/2007/pascal/eccs07_dresden/noble_denis/eccs07_noble_psb_01.ppt

61
Proteins
p Proteins are polypeptides
(strings of amino acid
residues)
p Represented using strings
of letters from an alphabet
of 20: AEGLV…WKKLAG
p Typical length 50…1000
residues
Urease enzyme from Helicobacter pylori

62
Amino acids
http://upload.wikimedia.org/wikipedia/commons/c/c5/Amino_acids_2.png

63
How DNA/RNA codes for protein?
p DNA alphabet contains four
letters but must specify
protein, or polypeptide
sequence of 20 letters.
p Dinucleotides are not
enough: 42 = 16 possible
dinucleotides
p Trinucleotides (triplets)
allow 43 = 64 possible
trinucleotides
p Triplets are also called
codons

64
How DNA/RNA codes for protein?
p Three of the possible
triplets specify ”stop
translation”
p Translation usually starts
at triplet AUG (this codes
for methionine)
p Most amino acids may be
specified by more than
triplet
p How to find a gene? Look
for start and stop codons
(not that easy though)

65
Proteins: Workhorses of the Cell
p 20 different amino acids
n different chemical properties cause the protein chains to fold
up into specific three-dimensional structures that define their
particular functions in the cell.
p Proteins do all essential work for the cell
n build cellular structures
n digest nutrients
n execute metabolic functions
n mediate information flow within a cell and among cellular
communities.
p Proteins work together with other proteins or nucleic acids
as "molecular machines"
n structures that fit together and function in highly specific, lock-
and-key ways.
Lecture 8: Proteomics

66
Genes
p “A gene is a union of genomic sequences encoding a
coherent set of potentially overlapping functional products”
--Gerstein et al.
p A DNA segment whose information is expressed either as
an RNA molecule or protein
5’ 3’
3’ 5’
… a t g a g t g g a …
… t a c t c a c c t …
augagugga ...
(transcription)
(translation)
MSG …
(folding)
http://fold.it

67
FoldIt: Protein folding game
http://fold.it

68
Genes & alleles
p A gene can have different variants
p The variants of the same gene are called
alleles
5’
3’
augagugga ...
MSG …
5’
3’
… a t g a g t c g a …
… t a c t c a g c t …
augagucga ...
MSR …

69
Genes can be found on both strands
3’
5’
5’
3’

70
Exons and introns & splicing
3’
5’
5’
3’
Introns are removed from RNA after transcription
Exons
Exons are joined:
This process is called splicing

71
Alternative splicing
A 3’
5’
5’
3’
B C
Different splice variants may be generated
A B C
B C
A C
…

72
Where does the variation in genomes come
from?
p Prokaryotes are typically
haploid: they have a single
(circular) chromosome
p DNA is usually inherited
vertically (parent to
daughter)
p Inheritance is clonal
n Descendants are faithful
copies of an ancestral DNA
n Variation is introduced via
mutations, transposable
elements, and horizontal
transfer of DNA
Chromosome map of S. dysenteriae, the nine rings
describe different properties of the genome
http://www.mgc.ac.cn/ShiBASE/circular_Sd197.htm

73
Causes of variation
p Mistakes in DNA replication
p Environmental agents (radiation, chemical
agents)
p Transposable elements (transposons)
n A part of DNA is moved or copied to another location in
genome
p Horizontal transfer of DNA
n Organism obtains genetic material from another
organism that is not its parent
n Utilized in genetic engineering

74
Biological string manipulation
p Point mutation: substitution of a base
n …ACGGCT… => …ACGCCT…
p Deletion: removal of one or more contiguous
bases (substring)
n …TTGATCA… => …TTTCA…
p Insertion: insertion of a substring
n …GGCTAG… => …GGTCAACTAG…
Lecture: Sequence alignment
Lecture: Genome rearrangements

75
Meiosis
p Sexual organisms are usually
diploid
n Germline cells (gametes)
contain N chromosomes
n Somatic (body) cells have 2N
chromosomes
p Meiosis: reduction of
chromosome number from
2N to N during reproductive
cycle
n One chromosome doubling is
followed by two cell divisions
Major events in meiosis
http://en.wikipedia.org/wiki/Meiosis
http://www.ncbi.nlm.nih.gov/About/Primer

76
Recombination and variation
p Recap: Allele is a viable DNA
coding occupying a given locus
(position in the genome)
p In recombination, alleles from
parents become suffled in
offspring individuals via
chromosomal crossover over
p Allele combinations in
offspring are usually different
from combinations found in
parents
p Recombination errors lead into
additional variations
Chromosomal crossover as described by
T. H. Morgan in 1916

77
Mitosis
http://en.wikipedia.org/wiki/Image:Major_events_in_mitosis.svg
p Mitosis: growth and development of the organism
n One chromosome doubling is followed by one cell
division

78
Recombination frequency and linked genes
p Genetic marker: some DNA sequence of interest
(e.g., gene or a part of a gene)
p Recombination is more likely to separate two
distant markers than two close ones
p Linked markers: ”tend” to be inherited together
p Marker distances measured in centimorgans: 1
centimorgan corresponds to 1% chance that two
markers are separated in recombination

79
Biological databases
p Exponential growth of
biological data
n New measurement
techniques
n Before we are able to use
the data, we need to store
it efficiently -> biological
databases
n Published data is
submitted to databases
p General vs specialised
databases
p This topic is discussed
extensively in Practical
course in biodatabases (III
period)

80
10 most important biodatabases… according
to ”Bioinformatics for dummies”
p GenBank/DDJB/EMBL www.ncbi.nlm.nih.gov Nucleotide sequences
p Ensembl www.ensembl.org Human/mouse genome
p PubMed www.ncbi.nlm.nih.gov Literature references
p NR www.ncbi.nlm.nih.gov Protein sequences
p UniProt www.expasy.org Protein sequences
p InterPro www.ebi.ac.uk Protein domains
p OMIM www.ncbi.nlm.nih.gov Genetic diseases
p Enzymes www.expasy.org Enzymes
p PDB www.rcsb.org/pdb/ Protein structures
p KEGG www.genome.ad.jp Metabolic pathways
Sophia Kossida, Introduction to Bioinformatics, Summer 2008

81
FASTA format
p A simple format for DNA and protein sequence
data is FASTA
>Hepatitis delta virus, complete genome
atgagccaagttccgaacaaggattcgcggggaggatagatcagcgcccgagaggggtga
gtcggtaaagagcattggaacgtcggagatacaactcccaagaaggaaaaaagagaaagc
aagaagcggatgaatttccccataacgccagtgaaactctaggaaggggaaagagggaag
gtggaagagaaggaggcgggcctcccgatccgaggggcccggcggccaagtttggaggac
actccggcccgaagggttgagagtaccccagagggaggaagccacacggagtagaacaga
gaaatcacctccagaggaccccttcagcgaacagagagcgcatcgcgagagggagtagac
catagcgataggaggggatgctaggagttgggggagaccgaagcgaggaggaaagcaaag
agagcagcggggctagcaggtgggtgttccgccccccgagaggggacgagtgaggcttat
cccggggaactcgacttatcgtccccacatagcagactcccggaccccctttcaaagtga
…
Header line,
begins with >

Introduction to
Bioinformatics
Biological words

83
Recap
p DNA codes information with alphabet of 4
letters: A, C, G, T
p In proteins, the alphabet size is 20
p DNA -> RNA -> Protein (genetic code)
n Three DNA bases (triplet, codon) code for one
amino acid
n Redundancy, start and stop codons

84
1 atgagccaag ttccgaacaa ggattcgcgg ggaggataga tcagcgcccg agaggggtga
61 gtcggtaaag agcattggaa cgtcggagat acaactccca agaaggaaaa aagagaaagc
121 aagaagcgga tgaatttccc cataacgcca gtgaaactct aggaagggga aagagggaag
181 gtggaagaga aggaggcggg cctcccgatc cgaggggccc ggcggccaag tttggaggac
241 actccggccc gaagggttga gagtacccca gagggaggaa gccacacgga gtagaacaga
301 gaaatcacct ccagaggacc ccttcagcga acagagagcg catcgcgaga gggagtagac
361 catagcgata ggaggggatg ctaggagttg ggggagaccg aagcgaggag gaaagcaaag
421 agagcagcgg ggctagcagg tgggtgttcc gccccccgag aggggacgag tgaggcttat
481 cccggggaac tcgacttatc gtccccacat agcagactcc cggaccccct ttcaaagtga
541 ccgagggggg tgactttgaa cattggggac cagtggagcc atgggatgct cctcccgatt
601 ccgcccaagc tccttccccc caagggtcgc ccaggaatgg cgggacccca ctctgcaggg
661 tccgcgttcc atcctttctt acctgatggc cggcatggtc ccagcctcct cgctggcgcc
721 ggctgggcaa cattccgagg ggaccgtccc ctcggtaatg gcgaatggga cccacaaatc
781 tctctagctt cccagagaga agcgagagaa aagtggctct cccttagcca tccgagtgga
841 cgtgcgtcct ccttcggatg cccaggtcgg accgcgagga ggtggagatg ccatgccgac
901 ccgaagagga aagaaggacg cgagacgcaa acctgcgagt ggaaacccgc tttattcact
961 ggggtcgaca actctgggga gaggagggag ggtcggctgg gaagagtata tcctatggga
1021 atccctggct tccccttatg tccagtccct ccccggtccg agtaaagggg gactccggga
1081 ctccttgcat gctggggacg aagccgcccc cgggcgctcc cctcgttcca ccttcgaggg
1141 ggttcacacc cccaacctgc gggccggcta ttcttctttc ccttctctcg tcttcctcgg
1201 tcaacctcct aagttcctct tcctcctcct tgctgaggtt ctttcccccc gccgatagct
1261 gctttctctt gttctcgagg gccttccttc gtcggtgatc ctgcctctcc ttgtcggtga
1321 atcctcccct ggaaggcctc ttcctaggtc cggagtctac ttccatctgg tccgttcggg
1381 ccctcttcgc cgggggagcc ccctctccat ccttatcttt ctttccgaga attcctttga
1441 tgtttcccag ccagggatgt tcatcctcaa gtttcttgat tttcttctta accttccgga
1501 ggtctctctc gagttcctct aacttctttc ttccgctcac ccactgctcg agaacctctt
1561 ctctcccccc gcggtttttc cttccttcgg gccggctcat cttcgactag aggcgacggt
1621 cctcagtact cttactcttt tctgtaaaga ggagactgct ggccctgtcg cccaagttcg
1681 ag
Given a DNA sequence, we might ask a number of questions
What sort of statistics should be used to describe the sequence?
What sort of organism did this sequence come from?
Does the description of this sequence differ from
the description of other DNA in the organism?
What sort of sequence is this? What does it do?

85
Biological words
p We can try to answer questions like these
by considering the words in a sequence
p A k-word (or a k-tuple) is a string of length
k drawn from some alphabet
p A DNA k-word is a string of length k that
consists of letters A, C, G, T
n 1-words: individual nucleotides (bases)
n 2-words: dinucleotides (AA, AC, AG, AT, CA, ...)
n 3-words: codons (AAA, AAC, …)
n 4-words and beyond

86
1-words: base composition
p Typically DNA exists as duplex molecule
(two complementary strands)
5’-GGATCGAAGCTAAGGGCT-3’
3’-CCTAGCTTCGATTCCCGA-5’
Top strand: 7 G, 3 C, 5 A, 3 T
Bottom strand: 3 G, 7 C, 3 A, 5 T
Duplex molecule: 10 G, 10 C, 8 A, 8 T
Base frequencies: 10/36 10/36 8/36 8/36
fr(G + C) = 20/36, fr(A + T) = 1 – fr(G + C) = 16/36
These are something
we can determine
experimentally.

87
G+C content
p fr(G + C), or G+C content is a simple
statistics for describing genomes
p Notice that one value is enough
characterise fr(A), fr(C), fr(G) and fr(T)
for duplex DNA
p Is G+C content (= base composition) able
to tell the difference between genomes of
different organisms?
n Simple computational experiment, if we have
the genome sequences under study
(-> exercises)

88
G+C content and genome sizes (in
megabasepairs, Mb) for various organisms
p Mycoplasma genitalium 31.6% 0.585
p Escherichia coli K-12 50.7% 4.693
p Pseudomonas aeruginosa PAO1 66.4% 6.264
p Pyrococcus abyssi 44.6% 1.765
p Thermoplasma volcanium 39.9% 1.585
p Caenorhabditis elegans 36% 97
p Arabidopsis thaliana 35% 125
p Homo sapiens 41% 3080

89
Base frequencies in duplex molecules
p Consider a DNA sequence generated randomly,
with probability of each letter being independent
of position in sequence
p You could expect to find a uniform distribution of
bases in genomes…
p This is not, however, the case in genomes,
especially in prokaryotes
n This phenomena is called GC skew
5’-...GGATCGAAGCTAAGGGCT...-3’
3’-...CCTAGCTTCGATTCCCGA...-5’

90
DNA replication fork
p When DNA is replicated,
the molecule takes the
replication fork form
p New complementary
DNA is synthesised at
both strands of the
”fork”
p New strand in 5’-3’
direction corresponding
to replication fork
movement is called
leading strand and the
other lagging strand
Leading strand
Lagging strand
Replication fork
Replication fork movement

91
DNA replication fork
p This process has
specific starting
points in genome
(origins of
replication)
p Observation:
Leading strands
have an excess of G
over C
p This can be
described by GC
skew statistics Lagging strand
Replication fork
Leading strand
Replication fork movement

92
GC skew
p GC skew is defined as (#G - #C) / (#G +
#C)
p It is calculated at successive positions in
intervals (windows) of specific width
5’-...GGATCGAAGCTAAGGGCT...-3’
3’-...CCTAGCTTCGATTCCCGA...-5’
(3 – 2) / (3 + 2) = 1/5
(4 – 2) / (4 + 2) = 1/3

93
p G-C content &
GC skew
statistics can be
displayed with a
circular genome
map
Chromosome map of S. dysenteriae, the nine rings
describe different properties of the genome
http://www.mgc.ac.cn/ShiBASE/circular_Sd197.htm
G-C content & GC skew
G+C content
GC skew
(10kb window size)

94
GC skew
p GC skew
often
changes sign
at origins
and termini
of replication
G+C content
GC skew
(10kb window size)
Nie et al., BMC Genomics, 2006

95
2-words: dinucleotides
p Let’s consider a sequence L1,L2,...,Ln
where each letter Li is drawn from the
DNA alphabet {A, C, G, T}
p We have 16 possible dinucleotides lili+1:
AA, AC, AG, ..., TG, TT.

96
i.i.d. model for nucleotides
p Assume that bases
n occur independently of each other
n bases at each position are identically
distributed
p Probability of the base A, C, G, T occuring
is pA, pC, pG, pT, respectively
n For example, we could use pA=pC=pG=pT=0.25
or estimate the values from known genome
data
p Probability of lili+1 is then PliPli+1
n For example, P(TG) = pT pG

97
2-words: is what we see surprising?
p We can test whether a sequence is ”unexpected”,
for example, with a 2 test
p Test statistic for a particular dinucleotide r1r2 is
2 = (O – E)2 / E where
n O is the observed number of dinucleotide r1r2
n E is the expected number of dinucleotide r1r2
n E = (n – 1)pr1pr2 under i.i.d. model
p Basic idea: high values of 2 indicate deviation
from the model
n Actual procedure is more detailed -> basic statistics
courses

98
Refining the i.i.d. model
p i.i.d. model describes some organisms well
(see Deonier’s book) but fails to
characterise many others
p We can refine the model by having the
DNA letter at some position depend on
letters at preceding positions
…TCGTGACGCCG ?
Sequence context to
consider

99
First-order Markov chains
p Lets assume that in sequence X the letter at
position t, Xt, depends only on the previous letter
Xt-1 (first-order markov chain)
p Probability of letter j occuring at position t given
Xt-1 = i: pij = P(Xt = j | Xt-1 = i)
p We consider homogeneous markov chains:
probability pij is independent of position t
…TCGTGACGCCG ?
Xt
Xt-1

100
Estimating pij
p We can estimate probabilities pij (”the probability
that j follows i”) from observed dinucleotide
frequencies
A C G T
A pAA pAC pAG pAT
C pCA pCC pCG pCT
G pGA pGC pGG pGT
T pTA pTC pTG pTT
Frequency
of dinucleotide AT
in sequence
…the values pAA, pAC, ..., pTG, pTT sum to 1
+ + + Base frequency
fr(C)

101
Estimating pij
p pij = P(Xt = j | Xt-1 = i) = P(Xt = j, Xt-1 = i)
P(Xt-1 = i)
Probability of transition i -> j
Dinucleotide frequency
Base frequency of nucleotide i,
fr(i)
A C G T
A 0.146 0.052 0.058 0.089
C 0.063 0.029 0.010 0.056
G 0.050 0.030 0.028 0.051
T 0.086 0.047 0.063 0.140
P(Xt = j, Xt-1 = i)
A C G T
A 0.423 0.151 0.168 0.258
C 0.399 0.184 0.063 0.354
G 0.314 0.189 0.176 0.321
T 0.258 0.138 0.187 0.415
P(Xt = j | Xt-1 = i)
0.052 / 0.345 0.151

102
Simulating a DNA sequence
p From a transition matrix, it is easy to generate a
DNA sequence of length n:
n First, choose the starting base randomly according to
the base frequency distribution
n Then, choose next base according to the distribution
P(xt | xt-1) until n bases have been chosen
A C G T
A 0.423 0.151 0.168 0.258
C 0.399 0.184 0.063 0.354
G 0.314 0.189 0.176 0.321
T 0.258 0.138 0.187 0.415
P(Xt = j | Xt-1 = i)
T T C T T C A
A
Look for R code in Deonier’s
book

103
ttcttcaaaataaggatagtgattcttattggcttaagggataacaatttagatcttttttcatgaatcatgtatgtcaacgttaaaagttgaactgcaataagttc
ttacacacgattgtttatctgcgtgcgaagcatttcactacatttgccgatgcagccaaaagtatttaacatttggtaaacaaattgacttaaatcgcgcacttaga
gtttgacgtttcatagttgatgcgtgtctaacaattacttttagttttttaaatgcgtttgtctacaatcattaatcagctctggaaaaacattaatgcatttaaac
cacaatggataattagttacttattttaaaattcacaaagtaattattcgaatagtgccctaagagagtactggggttaatggcaaagaaaattactgtagtgaaga
ttaagcctgttattatcacctgggtactctggtgaatgcacataagcaaatgctacttcagtgtcaaagcaaaaaaatttactgataggactaaaaaccctttattt
ttagaatttgtaaaaatgtgacctcttgcttataacatcatatttattgggtcgttctaggacactgtgattgccttctaactcttatttagcaaaaaattgtcata
gctttgaggtcagacaaacaagtgaatggaagacagaaaaagctcagcctagaattagcatgttttgagtggggaattacttggttaactaaagtgttcatgactgt
tcagcatatgattgttggtgagcactacaaagatagaagagttaaactaggtagtggtgatttcgctaacacagttttcatacaagttctattttctcaatggtttt
ggataagaaaacagcaaacaaatttagtattattttcctagtaaaaagcaaacatcaaggagaaattggaagctgcttgttcagtttgcattaaattaaaaatttat
ttgaagtattcgagcaatgttgacagtctgcgttcttcaaataagcagcaaatcccctcaaaattgggcaaaaacctaccctggcttctttttaaaaaaccaagaaa
agtcctatataagcaacaaatttcaaaccttttgttaaaaattctgctgctgaataaataggcattacagcaatgcaattaggtgcaaaaaaggccatcctctttct
ttttttgtacaattgttcaagcaactttgaatttgcagattttaacccactgtctatatgggacttcgaattaaattgactggtctgcatcacaaatttcaactgcc
caatgtaatcatattctagagtattaaaaatacaaaaagtacaattagttatgcccattggcctggcaatttatttactccactttccacgttttggggatatttta
acttgaatagttcacaatcaaaacataggaaggatctactgctaaaagcaaaagcgtattggaatgataaaaaactttgatgtttaaaaaactacaaccttaatgaa
ttaaagttgaaaaaatattcaaaaaaagaaattcagttcttggcgagtaatatttttgatgtttgagatcagggttacaaaataagtgcatgagattaactcttcaa
atataaactgatttaagtgtatttgctaataacattttcgaaaaggaatattatggtaagaattcataaaaatgtttaatactgatacaactttcttttatatcctc
catttggccagaatactgttgcacacaactaattggaaaaaaaatagaacgggtcaatctcagtgggaggagaagaaaaaagttggtgcaggaaatagtttctacta
acctggtataaaaacatcaagtaacattcaaattgcaaatgaaaactaaccgatctaagcattgattgatttttctcatgcctttcgcctagttttaataaacgcgc
cccaactctcatcttcggttcaaatgatctattgtatttatgcactaacgtgcttttatgttagcatttttcaccctgaagttccgagtcattggcgtcactcacaa
atgacattacaatttttctatgttttgttctgttgagtcaaagtgcatgcctacaattctttcttatatagaactagacaaaatagaaaaaggcacttttggagtct
gaatgtcccttagtttcaaaaaggaaattgttgaattttttgtggttagttaaattttgaacaaactagtatagtggtgacaaacgatcaccttgagtcggtgacta
taaaagaaaaaggagattaaaaatacctgcggtgccacattttttgttacgggcatttaaggtttgcatgtgttgagcaattgaaacctacaactcaataagtcatg
ttaagtcacttctttgaaaaaaaaaaagaccctttaagcaagctc
p Now we can quickly generate sequences of
arbitrary length...

104
aa 0.145 0.146
ac 0.050 0.052
ag 0.055 0.058
at 0.092 0.089
ca 0.065 0.063
cc 0.028 0.029
cg 0.011 0.010
ct 0.058 0.056
ga 0.048 0.050
gc 0.032 0.030
gg 0.029 0.028
gt 0.050 0.051
ta 0.084 0.086
tc 0.052 0.047
tg 0.064 0.063
tt 0.138 0.0140
Dinucleotide frequencies
Simulated Observed
n = 10000

105
ttcttcaaaataaggatagtgattcttattggcttaagggataacaatttagatcttttttcatgaatcatgtatgtcaacgttaaaagttgaactgcaataagttc
ttacacacgattgtttatctgcgtgcgaagcatttcactacatttgccgatgcagccaaaagtatttaacatttggtaaacaaattgacttaaatcgcgcacttaga
gtttgacgtttcatagttgatgcgtgtctaacaattacttttagttttttaaatgcgtttgtctacaatcattaatcagctctggaaaaacattaatgcatttaaac
cacaatggataattagttacttattttaaaattcacaaagtaattattcgaatagtgccctaagagagtactggggttaatggcaaagaaaattactgtagtgaaga
ttaagcctgttattatcacctgggtactctggtgaatgcacataagcaaatgctacttcagtgtcaaagcaaaaaaatttactgataggactaaaaaccctttattt
ttagaatttgtaaaaatgtgacctcttgcttataacatcatatttattgggtcgttctaggacactgtgattgccttctaactcttatttagcaaaaaattgtcata
gctttgaggtcagacaaacaagtgaatggaagacagaaaaagctcagcctagaattagcatgttttgagtggggaattacttggttaactaaagtgttcatgactgt
tcagcatatgattgttggtgagcactacaaagatagaagagttaaactaggtagtggtgatttcgctaacacagttttcatacaagttctattttctcaatggtttt
ggataagaaaacagcaaacaaatttagtattattttcctagtaaaaagcaaacatcaaggagaaattggaagctgcttgttcagtttgcattaaattaaaaatttat
ttgaagtattcgagcaatgttgacagtctgcgttcttcaaataagcagcaaatcccctcaaaattgggcaaaaacctaccctggcttctttttaaaaaaccaagaaa
agtcctatataagcaacaaatttcaaaccttttgttaaaaattctgctgctgaataaataggcattacagcaatgcaattaggtgcaaaaaaggccatcctctttct
ttttttgtacaattgttcaagcaactttgaatttgcagattttaacccactgtctatatgggacttcgaattaaattgactggtctgcatcacaaatttcaactgcc
caatgtaatcatattctagagtattaaaaatacaaaaagtacaattagttatgcccattggcctggcaatttatttactccactttccacgttttggggatatttta
acttgaatagttcacaatcaaaacataggaaggatctactgctaaaagcaaaagcgtattggaatgataaaaaactttgatgtttaaaaaactacaaccttaatgaa
ttaaagttgaaaaaatattcaaaaaaagaaattcagttcttggcgagtaatatttttgatgtttgagatcagggttacaaaataagtgcatgagattaactcttcaa
atataaactgatttaagtgtatttgctaataacattttcgaaaaggaatattatggtaagaattcataaaaatgtttaatactgatacaactttcttttatatcctc
catttggccagaatactgttgcacacaactaattggaaaaaaaatagaacgggtcaatctcagtgggaggagaagaaaaaagttggtgcaggaaatagtttctacta
acctggtataaaaacatcaagtaacattcaaattgcaaatgaaaactaaccgatctaagcattgattgatttttctcatgcctttcgcctagttttaataaacgcgc
cccaactctcatcttcggttcaaatgatctattgtatttatgcactaacgtgcttttatgttagcatttttcaccctgaagttccgagtcattggcgtcactcacaa
atgacattacaatttttctatgttttgttctgttgagtcaaagtgcatgcctacaattctttcttatatagaactagacaaaatagaaaaaggcacttttggagtct
gaatgtcccttagtttcaaaaaggaaattgttgaattttttgtggttagttaaattttgaacaaactagtatagtggtgacaaacgatcaccttgagtcggtgacta
taaaagaaaaaggagattaaaaatacctgcggtgccacattttttgttacgggcatttaaggtttgcatgtgttgagcaattgaaacctacaactcaataagtcatg
ttaagtcacttctttgaaaaaaaaaaagaccctttaagcaagctc
p The model is able to generate correct proportions
of 1- and 2-words in genomes...
p ...but fails with k=3 and beyond.

106
3-words: codons
p We can extend the previous method to 3-
words
p k=3 is an important case in study of DNA
sequences because of genetic code
5’ 3’
3’ 5’
a u g a g u g g a ...
M S G …

107
3-word probabilities
p Let’s again assume a sequence L of
independent bases
p Probability of 3-word r1r2r3 at position
i,i+1,i+2 in sequence L is
P(Li = r1, Li+1 = r2, Li+2 = r3) =
P(Li = r1)P(Li+1 = r2)P(Li+2 = r3)

108
3-words in Escherichia coli genome
AAA 108924 0.02348 0.01492
AAC 82582 0.01780 0.01541
AAG 63369 0.01366 0.01537
AAT 82995 0.01789 0.01490
ACA 58637 0.01264 0.01541
ACC 74897 0.01614 0.01591
ACG 73263 0.01579 0.01588
ACT 49865 0.01075 0.01539
AGA 56621 0.01220 0.01537
AGC 80860 0.01743 0.01588
AGG 50624 0.01091 0.01584
AGT 49772 0.01073 0.01536
ATA 63697 0.01373 0.01490
ATC 86486 0.01864 0.01539
ATG 76238 0.01643 0.01536
ATT 83398 0.01797 0.01489
CAA 76614 0.01651 0.01541
CAC 66751 0.01439 0.01591
CAG 104799 0.02259 0.01588
CAT 76985 0.01659 0.01539
CCA 86436 0.01863 0.01591
CCC 47775 0.01030 0.01643
CCG 87036 0.01876 0.01640
CCT 50426 0.01087 0.01589
CGA 70938 0.01529 0.01588
CGC 115695 0.02494 0.01640
CGG 86877 0.01872 0.01636
CGT 73160 0.01577 0.01586
CTA 26764 0.00577 0.01539
CTC 42733 0.00921 0.01589
CTG 102909 0.02218 0.01586
CTT 63655 0.01372 0.01537
Word Count Observed Expected Word Count Observed Expected

109
3-words in Escherichia coli genome
GAA 83494 0.01800 0.01537
GAC 54737 0.01180 0.01588
GAG 42465 0.00915 0.01584
GAT 86551 0.01865 0.01536
GCA 96028 0.02070 0.01588
GCC 92973 0.02004 0.01640
GCG 114632 0.02471 0.01636
GCT 80298 0.01731 0.01586
GGA 56197 0.01211 0.01584
GGC 92144 0.01986 0.01636
GGG 47495 0.01024 0.01632
GGT 74301 0.01601 0.01582
GTA 52672 0.01135 0.01536
GTC 54221 0.01169 0.01586
GTG 66117 0.01425 0.01582
GTT 82598 0.01780 0.01534
TAA 68838 0.01484 0.01490
TAC 52592 0.01134 0.01539
TAG 27243 0.00587 0.01536
TAT 63288 0.01364 0.01489
TCA 84048 0.01812 0.01539
TCC 56028 0.01208 0.01589
TCG 71739 0.01546 0.01586
TCT 55472 0.01196 0.01537
TGA 83491 0.01800 0.01536
TGC 95232 0.02053 0.01586
TGG 85141 0.01835 0.01582
TGT 58375 0.01258 0.01534
TTA 68828 0.01483 0.01489
TTC 83848 0.01807 0.01537
TTG 76975 0.01659 0.01534
TTT 109831 0.02367 0.01487
Word Count Observed Expected Word Count Observed Expected

110
2nd order Markov Chains
p Markov chains readily generalise to higher orders
p In 2nd order markov chain, position t depends on
positions t-1 and t-2
p Transition matrix:
p Probabilistic models for DNA and amino acid
sequences will be discussed in Biological
sequence analysis course (II period)
A C G T
AA
AC
AG
AT
CA
...

111
Codon Adaptation Index (CAI)
p Observation: cells prefer certain codons
over others in highly expressed genes
n Gene expression: DNA is transcribed into RNA
(and possibly translated into protein)
Phe TTT 0.493 0.551 0.291
TTC 0.507 0.449 0.709
Ala GCT 0.246 0.145 0.275
GCC 0.254 0.276 0.164
GCA 0.246 0.196 0.240
GCG 0.254 0.382 0.323
Asn AAT 0.493 0.409 0.172
AAC 0.507 0.591 0.828
Amino
acid Codon Predicted Gene class I Gene class II
Highly
expressed
Moderately
expressed
Codon frequencies for some genes in E. coli

112
p CAI is a statistic used to compare the
distribution of codons observed with the
preferred codons for highly expressed
genes

113
p Consider an amino acid sequence X = x1x2...xn
p Let pk be the probability that codon k is used in
highly expressed genes
p Let qk be the highest probability that a codon
coding for the same amino acid as codon k has
n For example, if codon k is ”GCC”, the
corresponding amino acid is Alanine (see genetic
code table; also GCT, GCA, GCG code for Alanine)
n Assume that pGCC = 0.164, pGCT = 0.275, pGCA =
0.240, pGCG = 0.323
n Now qGCC = qGCT = qGCA = qGCG = 0.323

114
p CAI is defined as
p CAI can be given also in log-odds form:
log(CAI) = (1/n) log(pk / qk)
CAI = ( pk / qk )
k=1
n 1/n
k=1
n

115
CAI: example with an E. coli gene
M A L T K A E M S E Y L …
ATG GCG CTT ACA AAA GCT GAA ATG TCA GAA TAT CTG
1.00 0.47 0.02 0.45 0.80 0.47 0.79 1.00 0.43 0.79 0.19 0.02
0.06 0.02 0.47 0.20 0.06 0.21 0.32 0.21 0.81 0.02
0.28 0.04 0.04 0.28 0.03 0.04
0.20 0.03 0.05 0.20 0.01 0.03
0.01 0.04 0.01
0.89 0.18 0.89
ATG GCT TTA ACT AAA GCT GAA ATG TCT GAA TAT TTA
GCC TTG ACC AAG GCC GAG TCC GAG TAC TTG
GCA CTT ACA GCA TCA CTT
GCG CTC ACG GCG TCG CTC
CTA AGT CTA
CTG AGC CTG
1.00 0.20 0.04 0.04 0.80 0.47 0.79 1.00 0.03 0.79 0.19 0.89…
1.00 0.47 0.89 0.47 0.80 0.47 0.79 1.00 0.43 0.79 0.81 0.89
1/n
qk
pk

116
CAI: properties
p CAI = 1.0 : each codon was the most frequently
used codon in highly expressed genes
p Log-odds used to avoid numerical problems
n What happens if you multiply many values <1.0
together?
p In a sample of E.coli genes, CAI ranged from 0.2
to 0.85
p CAI correlates with mRNA levels: can be used to
predict high expression levels

117
Biological words: summary
p Simple 1-, 2- and 3-word models can
describe interesting properties of DNA
sequences
n GC skew can identify DNA replication origins
n It can also reveal genome rearrangement
events and lateral transfer of DNA
n GC content can be used to locate genes:
human genes are comparably GC-rich
n CAI predicts high gene expression levels

118
Biological words: summary
p k=3 models can help to identify correct
reading frames
n Reading frame starts from a start codon and
stops in a stop codon
n Consider what happens to translation when a
single extra base is introduced in a reading
frame
p Also word models for k > 3 have their
uses

119
Next lecture
p Genome sequencing & assembly – where
do we get sequence data?

120
Note on programming languages
p Working with probability distributions is
straightforward with R, for example
n Deonier’s book contains many computational
examples
n You can use R in CS Linux systems
p Python works too!

121
#!/usr/bin/env python
import sys, random
n = int(sys.argv[1])
tm = {'a' : {'a' : 0.423, 'c' : 0.151, 'g' : 0.168, 't' : 0.258},
'c' : {'a' : 0.399, 'c' : 0.184, 'g' : 0.063, 't' : 0.354},
'g' : {'a' : 0.314, 'c' : 0.189, 'g' : 0.176, 't' : 0.321},
't' : {'a' : 0.258, 'c' : 0.138, 'g' : 0.187, 't' : 0.415}}
pi = {'a' : 0.345, 'c' : 0.158, 'g' : 0.159, 't' : 0.337}
def choose(dist):
r = random.random()
sum = 0.0
keys = dist.keys()
for k in keys:
sum += dist[k]
if sum > r:
return k
return keys[-1]
c = choose(pi)
for i in range(n - 1):
sys.stdout.write(c)
c = choose(tm[c])
sys.stdout.write(c)
sys.stdout.write("n")
Example Python code for generating
DNA sequences with first-order
Markov chains.
Function choose(), returns a key (here ’a’, ’c’, ’g’ or
’t’) of the dictionary ’dist’ chosen randomly
according to probabilities in dictionary values.
Choose the first letter, then choose
next letter according to P(xt | xt-1).
Transition matrix
tm and initial
distribution pi.
Initialisation: use packages ’sys’ and ’random’,
read sequence length from input.

Introduction to
Bioinformatics
Genome sequencing & assembly

123
Genome sequencing & assembly
p DNA sequencing
n How do we obtain DNA sequence information from
organisms?
p Genome assembly
n What is needed to put together DNA sequence
information from sequencing?
p First statement of sequence assembly problem
(according to G. Myers):
n Peltola, Söderlund, Tarhio, Ukkonen: Algorithms for
some string matching problems arising in molecular
genetics. Proc. 9th IFIP World Computer Congress, 1983

124
?
Recovery of shredded newspaper

125
DNA sequencing
p DNA sequencing: resolving a nucleotide
sequence (whole-genome or less)
p Many different methods developed
n Maxam-Gilbert method (1977)
n Sanger method (1977)
n High-throughput methods

126
Sanger sequencing: sequencing by
synthesis
p A sequencing technique developed by Fred
Sanger
p Also called dideoxy sequencing

127
http://en.wikipedia.org/wiki/DNA_polymerase
DNA polymerase
p A DNA polymerase is an
enzyme that catalyzes
DNA synthesis
p DNA polymerase needs
a primer
n Synthesis proceeds
always in 5’->3’ direction

128
Dideoxy sequencing
p In Sanger sequencing, chain-terminating
dideoxynucleoside triphosphates (ddXTPs)
are employed
n ddATP, ddCTP, ddGTP, ddTTP lack the 3’-OH
tail of dXTPs
p A mixture of dXTPs with small amount of
ddXTPs is given to DNA polymerase with
DNA template and primer
p ddXTPs are given fluorescent labels

129
Dideoxy sequencing
p When DNA polymerase encounters a
ddXTP, the synthesis cannot proceed
p The process yields copied sequences of
different lengths
p Each sequence is terminated by a labeled
ddXTP

130
Determining the sequence
p Sequences are sorted
according to length by
capillary
electrophoresis
p Fluorescent signals
corresponding to
labels are registered
p Base calling:
identifying which base
corresponds to each
position in a read
n Non-trivial problem!
Output sequences from
base calling are called reads

131
Reads are short!
p Modern Sanger sequencers can produce
quality reads up to ~750 bases1
n Instruments provide you with a quality file for
bases in reads, in addition to actual sequence
data
p Compare the read length against the size
of the human genome (2.9x109 bases)
p Reads have to be assembled!
1 Nature Methods - 5, 16 - 18 (2008)

132
Problems with sequencing
p Sanger sequencing error rate per base
varies from 1% to 3%1
p Repeats in DNA
n For example, ~300 base Alu sequence
repeated is over million times in human
genome
n Repeats occur in different scales
p What happens if repeat length is longer
than read length?
n We will get back to this problem later
1 Jones, Pevzner (2004)

133
Shortest superstring problem
p Find the shortest string that ”explains” the
reads
p Given a set of strings (reads), find a
shortest string that contains all of them

134
Example: Shortest superstring
Set of strings: {000, 001, 010, 011, 100, 101, 110, 111}
Concetenation of strings: 000001010011100101110111
010
110
011
000
Shortest superstring: 0001110100
001
111
101
100

135
Shortest superstrings: issues
p NP-complete problem: unlike to have an
efficient (exact) algorithm
p Reads may be from either strand of DNA
p Is the shortest string necessarily the
correct assembly?
p What about errors in reads?
p Low coverage -> gaps in assembly
n Coverage: average number of times each base
occurs in the set of reads (e.g., 5x coverage)

136
Sequence assembly and combination
locks
p What is common with sequence assembly
and opening keypad locks?

137
Whole-genome shotgun sequence
p Whole-genome shotgun sequence
assembly starts with a large sample of
genomic DNA
1. Sample is randomly partitioned into inserts of
length > 500 bases
2. Inserts are multiplied by cloning them into a
vector which is used to infect bacteria
3. DNA is collected from bacteria and sequenced
4. Reads are assembled

138
Assembly of reads with Overlap-Layout-
Consensus algorithm
p Overlap
n Finding potentially overlapping reads
p Layout
n Finding the order of reads along DNA
p Consensus (Multiple alignment)
n Deriving the DNA sequence from the layout
p Next, the method is described at a very
abstract level, skipping a lot of details
Kececioglu, J.D. and E.W. Myers. 1995. Combinatorial algorithms for
DNA sequence assembly. Algorithmica 13: 7-51.

139
Finding overlaps
p First, pairwise overlap
alignment of reads is
resolved
p Reads can be from
either DNA strand:
The reverse
complement r* of
each read r has to be
considered
acggagtcc
agtccgcgctt
5’ 3’
3’ 5’
r1
r2
r1: tgagt, r1
*: actca
r2: tccac, r2
*: gtgga

140
Example sequence to assemble
p 20 reads:
5’ – CAGCGCGCTGCGTGACGAGTCTGACAAAGACGGTATGCGCATCG
TGATTGAAGTGAAACGCGATGCGGTCGGTCGGTGAAGTTGTGCT - 3’
# Read Read*
1 CATCGTCA TCACGATG
2 CGGTGAAG CTTCACCG
3 TATGCGCA TGCGCATA
4 GACGAGTC GACTCGTC
5 CTGACAAA TTTGTCAG
6 ATGCGCAT ATGCGCAT
7 ATGCGGTC GACCGCAT
8 CTGCGTGA TCACGCAG
9 GCGTGACG CGTCACGC
10 GTCGGTGA TCACCGAC
# Read Read*
11 GGTCGGTG CACCGACC
12 ATCGTGAT ATCACGAT
13 GCGCTGCG CGCAGCGC
14 GCATCGTG CACGATGC
15 AGCGCGCT AGCGCGCT
16 GAAGTTGT ACAACTTC
17 AGTGAAAC GTTTCACT
18 ACGCGATG CATCGCGT
19 GCGCATCG CGATGCGC
20 AAGTGAAA TTTCACTT

141
Finding overlaps
p Overlap between two reads
can be found with a
dynamic programming
algorithm
n Errors can be taken into
account
p Dynamic programming will
be discussed more on next
lecture
p Overlap scores stored into
the overlap matrix
n Entries (i, j) below the
diagonal denote overlap of
read ri and rj
*
1 CATCGTCA
6 ATGCGCAT
12 ATCGTGAT
Overlap(1, 6) = 3
Overlap(1, 12) = 7
1
6 12
3 7

142
Finding layout & consensus
p Method extends the
assembly greedily by
choosing the best
overlaps
p Both orientations are
considered
p Sequence is extended
as far as possible
7* GACCGCAT
6=6* ATGCGCAT
14 GCATCGTG
1 CATCGTGA
12 ATCGTGAT
19 GCGCATCG
13* CGCAGCGC
---------------------
CGCATCGTGAT
Ambiguous bases
Consensus sequence

143
Finding layout & consensus
p We move on to next best
overlaps and extend the
sequence from there
p The method stops when
there are no more overlaps
to consider
p A number of contigs is
produced
p Contig stands for
contiguous sequence,
resulting from merging
reads
2 CGGTGAAG
10 GTCGGTGA
11 GGTCGGTG
7 ATGCGGTC
---------------------
ATGCGGTCGGTGAAG

144
Whole-genome shotgun sequencing:
summary
p Ordering of the reads is initially unknown
p Overlaps resolved by aligning the reads
p In a 3x109 bp genome with 500 bp reads and 5x
coverage, there are ~107 reads and ~107(107-1)/2
= ~5x1013 pairwise sequence comparisons
… …
Original genome sequence
Reads
Non-overlapping
read
Overlapping reads
=> Contig

145
Repeats in DNA and genome assembly
Pop, Salzberg, Shumway (2002)
Two instances of the same repeat

146
Repeats in DNA cause problems in
sequence assembly
p Recap: if repeat length exceeds read
length, we might not get the correct
assembly
p This is a problem especially in eukaryotes
n ~3.1% of genome consists of repeats in
Drosophila, ~45% in human
p Possible solutions
1. Increase read length – feasible?
2. Divide genome into smaller parts, with known
order, and sequence parts individually

147
”Divide and conquer” sequencing
approaches: BAC-by-BAC
Whole-genome shotgun sequencing
Divide-and-conquer
Genome
Genome
BAC library

148
BAC-by-BAC sequencing
p Each BAC (Bacterial Artificial
Chromosome) is about 150 kbp
p Covering the human genome requires
~30000 BACs
p BACs shotgun-sequenced separately
n Number of repeats in each BAC is
significantly smaller than in the whole
genome...
n ...needs much more manual work compared
to whole-genome shotgun sequencing

149
Hybrid method
p Divide-and-conquer and whole-genome
shotgun approaches can be combined
n Obtain high coverage from whole-genome
shotgun sequencing for short contigs
n Generate of a set of BAC contigs with low
coverage
n Use BAC contigs to ”bin” short contigs to
correct places
p This approach was used to sequence the
brown Norway rat genome in 2004

150
Paired end sequencing
p Paired end (or mate-pair) sequencing is
technique where
n both ends of an insert are sequenced
n For each insert, we get two reads
n We know the distance between reads, and that
they are in opposite orientation
n Typically read length < insert length
k
Read 1 Read 2

151
Paired end sequencing
p The key idea of paired end sequencing:
n Both reads from an insert are unlikely to be in repeat
regions
n If we know where the first read is, we know also
second’s location
p This technique helps to WGSS higher organisms
k
Read 1 Read 2
Repeat region

152
First whole-genome shotgun sequencing
project: Drosophila melanogaster
p Fruit fly is a common
model organism in
biological studies
p Whole-genome
assembly reported in
Eugene Myers, et al.,
A Whole-Genome
Assembly of
Drosophila, Science
24, 2000
p Genome size 120 Mbp
http://en.wikipedia.org/wiki/Drosophila_melanogaster

153
Sequencing of the Human Genome
p The (draft) human
genome was published
in 2001
p Two efforts:
n Human Genome Project
(public consortium)
n Celera (private
company)
p HGP: BAC-by-BAC
approach
p Celera: whole-genome
shotgun sequencing
HGP: Nature 15 February 2001
Vol 409 Number 6822
Celera: Science 16 February 2001
Vol 291, Issue 5507

154
Genome assembly software
p phrap (Phil’s revised assembly program)
p AMOS (A Modular, Open-Source whole-
genome assembler)
p CAP3 / PCAP
p TIGR assembler

155
Next generation sequencing techniques
p Sanger sequencing is the prominent first-
generation sequencing method
p Many new sequencing methods are
emerging
p See Lars Paulin’s slides (course web page)
for details

156
Next-gen sequencing: 454
p Genome Sequencer FLX (454 Life Science
/ Roche)
n >100 Mb / 7.5 h run
n Read length 250-300 bp
n >99.5% accuracy / base in a single run
n >99.99% accuracy / base in consensus

157
Next-gen sequencing: Illumina Solexa
p Illumina / Solexa Genome Analyzer
n Read length 35 - 50 bp
n 1-2 Gb / 3-6 day run
n > 98.5% accuracy / base in a single run
n 99.99% accuracy / consensus with 3x
coverage

158
Next-gen sequencing: SOLiD
p SOLiD
n Read length 25-30 bp
n 1-2 Gb / 5-10 day run
n >99.94% accuracy / base
n >99.999% accuracy / consensus with 15x
coverage

159
Next-gen sequencing: Helicos
p Helicos: Single Molecule Sequencer
n No amplification of sequences needed
n Read length up to 55 bp
p Accuracy does not decrease when read length is
increased
p Instead, throughput goes down
n 25-90 Mb / h
n >2 Gb / day

160
Next-gen sequencing: Pacific
Biosciences
p Pacific Biosciences
n Single-Molecule Real-Time (SMRT) DNA
sequencing technology
n Read length “thousands of nucleotides”
p Should overcome most problems with repeats
n Throughput estimate: 100 Gb / hour
n First instruments in 2010?

161
Exercise groups
p 1st group: Tuesdays 16.15-18.00 C221
p 2nd group: Wednesday 14.15-16.00 B120
p You can choose freely which group you
want to attend to
p Send exercise notes before the 1st group
starts (Tue 16.15), even if you go to the
2nd group

Introduction to
Bioinformatics
Lecture 3:
Sequence alignment

163
Sequence alignment
p The biological problem
p Global alignment
p Local alignment
p Multiple alignment

164
Background: comparative genomics
p Basic question in biology: what properties
are shared among organisms?
p Genome sequencing allows comparison of
organisms at DNA and protein levels
p Comparisons can be used to
n Find evolutionary relationships between
organisms
n Identify functionally conserved sequences
n Identify corresponding genes in human and
model organisms: develop models for human
diseases

165
Homologs
p Two genes (sequences in
general) gB and gC
evolved from the same
ancestor gene gA are
called homologs
p Homologs usually exhibit
conserved functions
p Close evolutionary
relationship => expect a
high number of homologs
gB = agtgccgttaaagttgtacgtc
gC = ctgactgtttgtggttc
gA = agtgtccgttaagtgcgttc

166
p We expect homologs to be ”similar” to each other
p Intuitively, similarity of two sequences refers to
the degree of match between corresponding
positions in sequence
p What about sequences that differ in length?
Sequence similarity
agtgccgttaaagttgtacgtc
ctgactgtttgtggttc

167
Similarity vs homology
p Sequence similarity is not sequence
homology
n If the two sequences gB and gC have accumulated
enough mutations, the similarity between them is likely
to be low
Homology is more difficult to detect over greater
evolutionary distances.
0 agtgtccgttaagtgcgttc
1 agtgtccgttatagtgcgttc
2 agtgtccgcttatagtgcgttc
4 agtgtccgcttaagggcgttc
8 agtgtccgcttcaaggggcgt
16 gggccgttcatgggggt
32 gcagggcgtcactgagggct
64 acagtccgttcgggctattg
128 cagagcactaccgc
256 cacgagtaagatatagct
512 taatcgtgata
1024 acccttatctacttcctggagtt
2048 agcgacctgcccaa
4096 caaac
#mutations #mutations

168
Similarity vs homology (2)
p Sequence similarity can occur by chance
n Similarity does not imply homology
p Consider comparing two short sequences
against each other

169
Orthologs and paralogs
p We distinguish between two types of homology
n Orthologs: homologs from two different species,
separated by a speciation event
n Paralogs: homologs within a species, separated by a
gene duplication event
gA
gB gC
Organism B Organism C
gA
gA gA’
gB gC
Organism A
Gene duplication event
Orthologs
Paralogs

170
Orthologs and paralogs (2)
p Orthologs typically retain the original function
p In paralogs, one copy is free to mutate and
acquire new function (no selective pressure)
gA
gB gC
Organism B Organism C
gA
gA gA’
gB gC
Organism A

171
Paralogy example: hemoglobin
p Hemoglobin is a protein
complex which transports
oxygen
p In humans, hemoglobin
consists of four protein
subunits and four non-
protein heme groups
Hemoglobin A,
www.rcsb.org/pdb/explore.do?structureId=1GZX
Sickle cell diseases
are caused by mutations
in hemoglobin genes
http://en.wikipedia.org/wiki/Image:Sicklecells.jpg

172
p In adults, three types are
normally present
n Hemoglobin A: 2 alpha and
2 beta subunits
n Hemoglobin A2: 2 alpha
and 2 delta subunits
n Hemoglobin F: 2 alpha and
2 gamma subunits
p Each type of subunit
(alpha, beta, gamma,
delta) is encoded by a
separate gene
Hemoglobin A,

173
p The subunit genes are
paralogs of each other, i.e.,
they have a common ancestor
gene
p Exercise: hemoglobin human
paralogs in NCBI sequence
databases
http://www.ncbi.nlm.nih.gov/sites/entre
z?db=Nucleotide
n Find human hemoglobin alpha, beta,
gamma and delta
n Compare sequences
Hemoglobin A,

174
Orthology example: insulin
p The genes coding for insulin in human
(Homo sapiens) and mouse (Mus
musculus) are orthologs:
n They have a common ancestor gene in the
ancestor species of human and mouse
n Exercise: find insulin orthologs from human
and mouse in NCBI sequence databases

175
Sequence alignment
p Alignment specifies which positions in two
sequences match
acgtctag
|||||
-actctag
5 matches
2 mismatches
1 not aligned
acgtctag
||
actctag-
2 matches
5 mismatches
1 not aligned
acgtctag
|| |||||
ac-tctag
7 matches
0 mismatches
1 not aligned

176
Sequence alignment
p Maximum alignment length is the total length of
the two sequences
acgtctag-------
--------actctag
0 matches
0 mismatches
15 not aligned
-------acgtctag
actctag--------
0 matches
0 mismatches
15 not aligned

177
Mutations: Insertions, deletions and
substitutions
p Insertions and/or deletions are called
indels
n We can’t tell whether the ancestor sequence
had a base or not at indel position!
acgtctag
|||||
-actctag
Indel: insertion or
deletion of a base
with respect to the
ancestor sequence
Mismatch: substitution
(point mutation) of
a single base

178
Problems
p What sorts of alignments should be considered?
p How to score alignments?
p How to find optimal or good scoring alignments?
p How to evaluate the statistical significance of
scores?
In this course, we discuss each of these problems
briefly.

179
Sequence Alignment (chapter 6)
p Global alignment
p Local alignment

180
Global alignment
p Problem: find optimal scoring alignment between
two sequences (Needleman & Wunsch 1970)
p Every position in both sequences is included in
the alignment
p We give score for each position in alignment
n Identity (match) +1
n Substitution (mismatch) -µ
n Indel -
p Total score: sum of position scores

181
Scoring: Toy example
p Consider two sequences
with characters drawn
from the English
language alphabet:
WHAT, WHY
WHAT
||
WH-Y
S(WHAT/WH-Y) = 1 + 1 – – µ
WHAT
-WHY
S(WHAT/-WHY) = – – µ – µ – µ

182
Dynamic programming
p How to find the optimal alignment?
p We use previous solutions for optimal
alignments of smaller subsequences
p This general approach is known as
dynamic programming

183
Introduction to dynamic programming:
the money change problem
p Suppose you buy a pen for 4.23€ and pay for
with a 5€ note
p You get 77 cents in change – what coins is the
cashier going to give you if he or she tries to
minimise the number of coins?
p The usual algorithm: start with largest coin
(denominator), proceed to smaller coins until no
change is left:
n 50, 20, 5 and 2 cents
p This greedy algorithm is incorrect, in the sense
that it does not always give you the correct
answer

184
The money change problem
p How else to compute the
change?
p We could consider all possible
ways to reduce the amount of
change
p Suppose we have 77 cents
change, and the following
coins: 50, 20, 5 cents
p We can compute the change
with recursion
n C(n) = min { C(n – 50) + 1,
C(n – 20) + 1, C(n – 5) + 1 }
p Figure shows the recursion
tree for the example
77
72
27 57
7 22 7 37 52 22 52 67
…
50 20 5
p Many values are computed more
than once!
p This leads to a correct but very
inefficient algorithm

185
p We can speed the computation up by
solving the change problem for all i n
n Example: solve the problem for 9 cents with
available coins being 1, 2 and 5 cents
p Solve the problem in steps, first for 1
cent, then 2 cents, and so on
p In each step, utilise the solutions from the
previous steps

186
0 1 2 3 4 5 6 7 8 9
Amount of
change left
p Algorithm runs in time proportional to Md, where M is the
amount of change and d is the number of coin types
p The same technique of storing solutions of subproblems can
be utilised in aligning sequences
Number of
coins used
0 1 1 2 2 1 2 2 3 3

187
Representing alignments and scores
Y
H
W
-
T
A
H
W
-
WHAT
||
WH-Y
Alignments can be
represented in the
following tabular form.
Each alignment
corresponds to a path
through the table.

188
Y
H
W
-
T
A
H
W
-
WHAT---
----WHY
WH-AT
||
WHY--

189
Y
H
W
-
T
A
H
W
-
WHAT
||
WH-Y
Global alignment
score S3,4 = 2- -µ
2- -µ
2-
2
1
0

190
Filling the alignment matrix
Y
H
W
-
T
A
H
W
-
Case 1
Case 2
Case 3
Consider the alignment process
at shaded square.
Case 1. Align H against H
(match)
Case 2. Align H in WHY against
– (indel) in WHAT
Case 3. Align H in WHAT
against – (indel) in WHY

191
Filling the alignment matrix (2)
Y
H
W
-
T
A
H
W
-
Case 1
Case 2
Case 3
Scoring the alternatives.
Case 1. S2,2 = S1,1 + s(2, 2)
Case 2. S2,2 = S1,2
Case 3. S2,2 = S2,1
s(i, j) = 1 for matching positions,
s(i, j) = - µ for substitutions.
Choose the case (path) that
yields the maximum score.
Keep track of path choices.

192
Global alignment: formal
development
A = a1a2a3…an,
B = b1b2b3…bm
a3
a2
a1
-
b4
b3
b2
b1
-
3
2
1
0
4
3
2
1
0
b1 b2 b3 b4 -
- -
a1 a2 a3
l Any alignment can be written
as a unique path through the
matrix
l Score for aligning A and B up
to positions i and j:
Si,j = S(a1a2a3…ai, b1b2b3…bj)

193
Scoring partial alignments
p Alignment of A = a1a2a3…ai with B = b1b2b3…bj
can be end in three possible ways
n Case 1: (a1a2…ai-1) ai
(b1b2…bj-1) bj
(b1b2…bj) -
n Case 3: (a1a2…ai) –
(b1b2…bj-1) bj

194
Scoring alignments
p Scores for each case:
(b1b2…bj-1) bj
(b1b2…bj) –
n Case 3: (a1a2…ai) –
(b1b2…bj-1) bj
s(ai, bj) = { -µ otherwise
+1 if ai = bj
s(ai, -) = s(-, bj) = -

195
Scoring alignments (2)
p First row and first column
correspond to initial alignment
against indels:
S(i, 0) = -i
S(0, j) = -j
p Optimal global alignment score
S(A, B) = Sn,m
a3
a2
a1
-
b4
b3
b2
b1
-
-3
3
-2
2
-
1
-4
-3
-2
-
0
0
4
3
2
1
0

196
Algorithm for global alignment
Input sequences A, B, n = |A|, m = |B|
Set Si,0 := - i for all i
Set S0,j := - j for all j
for i := 1 to n
for j := 1 to m
Si,j := max{Si-1,j – , Si-1,j-1 + s(ai,bj), Si,j-1 – }
end
end
Algorithm takes O(nm) time

197
Global alignment: example
?
-10
T
-8
G
-6
C
-4
T
-2
A
-10
-8
-6
-4
-2
0
-
G
T
G
G
T
-
µ = 1
= 2

198
Global alignment: example
?
-10
T
-8
G
-6
C
-4
T
-2
A
-10
-8
-6
-4
-2
0
-
G
T
G
G
T
-
µ = 1
= 2 -1 -3

199
Global alignment: example (2)
-2
0
-3
-4
-7
-10
T
-4
-3
-1
-2
-5
-8
G
-5
-5
-3
-2
-3
-6
C
-6
-4
-4
-2
-1
-4
T
-9
-7
-5
-3
-1
-2
A
-10
-8
-6
-4
-2
0
-
G
T
G
G
T
-
µ = 1
= 2
ATCGT-
| ||
-TGGTG

200
p Global alignment
p Local alignment

201
Local alignment: rationale
p Otherwise dissimilar proteins may have local regions of
similarity
-> Proteins may share a function
Human bone
morphogenic protein
receptor type II
precursor (left) has a
300 aa region that
resembles 291 aa
region in TGF-
receptor (right).
The shared function
here is protein kinase.

202
Local alignment: rationale
p Global alignment would be inadequate
p Problem: find the highest scoring local alignment
between two sequences
p Previous algorithm with minor modifications solves this
problem (Smith & Waterman 1981)
A
B
Regions of
similarity

203
From global to local alignment
p Modifications to the global alignment
algorithm
n Look for the highest-scoring path in the
alignment matrix (not necessarily through the
matrix), or in other words:
n Allow preceding and trailing indels without
penalty

204
Scoring local alignments
A = a1a2a3…an, B = b1b2b3…bm
Let I and J be intervals (substrings) of A and B, respectively:
Best local alignment score:
where S(I, J) is the alignment score for substrings I and J.

205
Allowing preceding and trailing
indels
p First row and column
initialised to zero:
Mi,0 = M0,j = 0
a3
a2
a1
-
b4
b3
b2
b1
-
0
3
0
2
0
1
0
0
0
0
0
0
4
3
2
1
0
b1 b2 b3
- - a1

206
Recursion for local alignment
p Mi,j = max {
Mi-1,j-1 + s(ai, bi),
Mi-1,j – ,
Mi,j-1 – ,
0
}
0
2
0
0
1
0
T
1
0
1
1
0
0
G
0
0
0
0
0
0
C
0
1
0
0
1
0
T
0
0
0
0
0
0
A
0
0
0
0
0
0
-
G
T
G
G
T
-
Allow alignment to
start anywhere in
sequences

207
Finding best local alignment
p Optimal score is the highest
value in the matrix
= maxi,j Mi,j
p Best local alignment can be
found by backtracking from the
highest value in M
p What is the best local
alignment in this example?
0
2
0
0
1
0
T
1
0
1
1
0
0
G
0
0
0
0
0
0
C
0
1
0
0
1
0
T
0
0
0
0
0
0
A
0
0
0
0
0
0
-
G
T
G
G
T
-

208
Local alignment: example
0
G
8
0
G
7
0
A
6
0
A
5
0
T
4
0
C
3
0
C
2
0
A
1
0
0
0
0
0
0
0
0
0
0
0
-
0
A
C
T
A
A
C
T
C
G
G
-
10
9
8
7
6
5
4
3
2
1
0
Mi,j = max {
Mi-1,j-1 + s(ai,
bi),
Mi-1,j ,
Mi,j-1 ,
0
}
0
Scoring (for example)
Match: +2
Mismatch: -1
Indel: -2

209
0
G
8
0
G
7
0
A
6
0
A
5
0
T
4
0
C
3
0
C
2
0
A
1
0
0
0
0
0
0
0
0
0
0
0
-
0
A
C
T
A
A
C
T
C
G
G
-
10
9
8
7
6
5
4
3
2
1
0
Mi,j = max {
Mi-1,j-1 + s(ai,
bi),
Mi-1,j ,
Mi,j-1 ,
0
}
0 0 0 0 0 2
Match: +2
Mismatch: -1
Indel: -2

210
C T – A A
C T C A A
Match: +2
Mismatch: -1
Indel: -2
Optimal local
alignment:
2
4
3
2
1
0
0
2
4
2
0
G
8
1
3
5
4
3
0
0
0
2
2
0
G
7
3
2
4
6
5
1
0
0
0
0
0
A
6
3
1
1
3
4
3
2
0
0
0
0
A
5
2
1
2
0
1
2
4
0
0
0
0
T
4
1
3
0
0
1
2
1
2
0
0
0
C
3
0
2
1
1
0
2
0
2
0
0
0
C
2
2
0
0
2
2
0
0
0
0
0
0
A
1
0
0
0
0
0
0
0
0
0
0
0
-
0
A
C
T
A
A
C
T
C
G
G
-
10
9
8
7
6
5
4
3
2
1
0

211
Multiple optimal alignments
Non-optimal, good-scoring alignments
2
4
3
2
1
0
0
2
4
2
0
G
8
1
3
5
4
3
0
0
0
2
2
0
G
7
3
2
4
6
5
1
0
0
0
0
0
A
6
3
1
1
3
4
3
2
0
0
0
0
A
5
2
1
2
0
1
2
4
0
0
0
0
T
4
1
3
0
0
1
2
1
2
0
0
0
C
3
0
2
1
1
0
2
0
2
0
0
0
C
2
2
0
0
2
2
0
0
0
0
0
0
A
1
0
0
0
0
0
0
0
0
0
0
0
-
0
A
C
T
A
A
C
T
C
G
G
-
10
9
8
7
6
5
4
3
2
1
0
How can you find
1. Optimal
alignments if
more than one
exist?
2. Non-optimal,
good-scoring
alignments?

212
Overlap alignment
p Overlap matrix used by Overlap-Layout-
Consensus algorithm can be computed with
dynamic programming
p Initialization: Oi,0 = O0,j = 0 for all i, j
p Recursion:
Oi,j = max {
Oi-1,j-1 + s(ai, bi),
Oi-1,j – ,
Oi,j-1 – ,
}
Best overlap: maximum value from rightmost
column and bottom row

213
Non-uniform mismatch penalties
p We used uniform penalty for mismatches:
s(’A’, ’C’) = s(’A’, ’G’) = … = s(’G’, ’T’) = µ
p Transition mutations (A->G, G->A, C->T, T->C) are
approximately twice as frequent than transversions (A->T,
T->A, A->C, G->T)
n use non-uniform mismatch
penalties collected into a
substitution matrix
1
-1
-0.5
-1
T
-1
1
-1
-0.5
G
-0.5
-1
1
-1
C
-1
-0.5
-1
1
A
T
G
C
A

214
Gaps in alignment
p Gap is a succession of indels in alignment
p Previous model scored a length k gap as
w(k) = -k
p Replication processes may produce longer
stretches of insertions or deletions
n In coding regions, insertions or deletions of
codons may preserve functionality
C T – - - A A
C T C G C A A

215
Gap open and extension penalties (2)
p We can design a score that allows the
penalty opening gap to be larger than
extending the gap:
w(k) = - – (k – 1)
p Gap open cost , Gap extension cost
p Alignment algorithms can be extended to
use w(k) (not discussed on this course)

216
Amino acid sequences
p We have discussed mainly DNA sequences
p Amino acid sequences can be aligned as
well
p However, the design of the substitution
matrix is more involved because of the
larger alphabet
p More on the topic in the course Biological
sequence analysis

217
Demonstration of the EBI web site
p European Bioinformatics Institute (EBI)
offers many biological databases and
bioinformatics tools at
http://www.ebi.ac.uk/
n Sequence alignment: Tools -> Sequence
Analysis -> Align

218
p Global alignment
p Local alignment

219
Multiple alignment
p Consider a set of n sequences
on the right
n Orthologous sequences from
different organisms
n Paralogs from multiple
duplications
p How can we study
relationships between these
sequences?
aggcgagctgcgagtgcta
cgttagattgacgctgac
ttccggctgcgac
gacacggcgaacgga
agtgtgcccgacgagcgaggac
gcgggctgtgagcgcta
aagcggcctgtgtgcccta
atgctgctgccagtgta
agtcgagccccgagtgc
agtccgagtcc
actcggtgc

220
Optimal alignment of three
sequences
p Alignment of A = a1a2…ai and B = b1b2…bj can
end either in (-, bj), (ai, bj) or (ai, -)
p 22 – 1 = 3 alternatives
p Alignment of A, B and C = c1c2…ck can end in 23 –
1 ways: (ai, -, -), (-, bj, -), (-, -, ck), (-, bj, ck),
(ai, -, ck), (ai, bj, -) or (ai, bj, ck)
p Solve the recursion using three-dimensional
dynamic programming matrix: O(n3) time and
space
p Generalizes to n sequences but impractical with
even a moderate number of sequences

221
Multiple alignment in practice
p In practice, real-world multiple alignment
problems are usually solved with heuristics
p Progressive multiple alignment
n Choose two sequences and align them
n Choose third sequence w.r.t. two previous sequences
and align the third against them
n Repeat until all sequences have been aligned
n Different options how to choose sequences and score
alignments
n Note the similarity to Overlap-Layout-Consensus

222
Multiple alignment in practice
p Profile-based progressive multiple
alignment: CLUSTALW
n Construct a distance matrix of all pairs of
sequences using dynamic programming
n Progressively align pairs in order of decreasing
similarity
n CLUSTALW uses various heuristics to
contribute to accuracy

223
Additional material
p R. Durbin, S. Eddy, A. Krogh, G.
Mitchison: Biological sequence analysis
p N. C. Jones, P. A. Pevzner: An introduction
to bioinformatics algorithms
p Course Biological sequence analysis in
period II, 2008

224
Rapid alignment methods: FASTA and
BLAST
p Search strategies
p FASTA
p BLAST

225
The biological problem
p Global and local
alignment algoritms are
slow in practice
p Consider the scenario of
aligning a query
sequence against a large
database of sequences
n New sequence with
unknown function n NCBI GenBank size in January
2007 was 65 369 091 950
bases (61 132 599 sequences)
n Feb 2008: 85 759 586 764
bases (82 853 685 sequences)

226
Problem with large amount of sequences
p Exponential growth in both number and
total length of sequences
p Possible solution: Compare against model
organisms only
p With large amount of sequences, chances
are that matches occur by random
n Need for statistical analysis

227
BLAST
p Search strategies
p FASTA
p BLAST

228
FASTA
p FASTA is a multistep algorithm for sequence
alignment (Wilbur and Lipman, 1983)
p The sequence file format used by the FASTA
software is widely used by other sequence
analysis software
p Main idea:
n Choose regions of the two sequences (query and
database) that look promising (have some degree of
similarity)
n Compute local alignment using dynamic programming in
these regions

229
FASTA outline
p FASTA algorithm has five steps:
n 1. Identify common k-words between I and J
n 2. Score diagonals with k-word matches,
identify 10 best diagonals
n 3. Rescore initial regions with a substitution
score matrix
n 4. Join initial regions using gaps, penalise for
gaps
n 5. Perform dynamic programming to find final
alignments

230
Search strategies
p How to speed up the computation?
n Find ways to limit the number of pairwise
comparisons
p Compare the sequences at word level to
find out common words
n Word means here a k-tuple (or a k-word), a
substring of length k

231
Analyzing the word content
p Example query string I: TGATGATGAAGACATCAG
p For k = 8, the set of k-words (substring of length
k) of I is
TGATGATG
GATGATGA
ATGATGAA
TGATGAAG
…
GACATCAG

232
Analyzing the word content
p There are n-k+1 k-words in a string of length n
p If at least one word of I is not found from
another string J, we know that I differs from J
p Need to consider statistical significance: I and J
might share words by chance only
p Let n=|I| and m=|J|

233
Word lists and comparison by content
p The k-words of I can be arranged into a table of
word occurences Lw(I)
p Consider the k-words when k=2 and
I=GCATCGGC:
GC, CA, AT, TC, CG, GG, GC
AT: 3
CA: 2
CG: 5
GC: 1, 7
GG: 6
TC: 4
Start indecies of k-word GC in I
Building Lw(I) takes O(n) time

234
Common k-words
p Number of common k-words in I and J can
be computed using Lw(I) and Lw(J)
p For each word w in I, there are |Lw(J)|
occurences in J
p Therefore I and J have
common words
p This can be computed in O(n + m + 4k)
time
n O(n + m) time to build the lists
n O(4k) time to calculate the sum (in DNA
strings)

235
Common k-words
p I = GCATCGGC
p J = CCATCGCCATCG
Lw(J)
AT: 3, 9
CA: 2, 8
CC: 1, 7
CG: 5, 11
GC: 6
TC: 4, 10
Lw(I)
AT: 3
CA: 2
CG: 5
GC: 1, 7
GG: 6
TC: 4
Common words
2
2
0
2
2
0
2
10 in total

236
Properties of the common word list
p Exact matches can be found using binary search
(e.g., where TCGT occurs in I?)
n O(log 4k) time
p For large k, the table size is too large to compute
the common word count in the previous fashion
p Instead, an approach based on merge sort can be
utilised (details skipped)
p The common k-word technique can be combined
with the local alignment algorithm to yield a rapid
alignment approach

237
FASTA outline
score matrix
gaps
alignments

238
Dot matrix comparisons
p Word matches in two sequences I and J can be
represented as a dot matrix
p Dot matrix element (i, j) has ”a dot”, if the word
starting at position i in I is identical to the word
starting at position j in J
p The dot matrix can be plotted for various k
i
j
I = … ATCGGATCA …
J = … TGGTGTCGC …
i
j

239
k=1 k=4
k=8 k=16
Dot matrix (k=1,4,8,16)
for two DNA sequences
X85973.1 (1875 bp)
Y11931.1 (2013 bp)

240
k=1 k=4
k=8 k=16
Dot matrix
(k=1,4,8,16) for two
protein sequences
CAB51201.1 (531 aa)
CAA72681.1 (588 aa)
Shading indicates
now the match score
according to a
score matrix
(Blosum62 here)

241
Computing diagonal sums
p We would like to find high scoring diagonals of the dot
matrix
p Lets index diagonals by the offset, l = i - j
C C A T C G C C A T C G
G *
C * *
A * *
T * *
C * *
G
G *
C
k=2
I
J
Diagonal l = i – j = -6

242
p As an example, lets compute diagonal sums for
I = GCATCGGC, J = CCATCGCCATCG, k = 2
p 1. Construct k-word list Lw(J)
p 2. Diagonal sums Sl are computed into a table,
indexed with the offset and initialised to zero
l -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Sl 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

243
p 3. Go through k-words of I, look for matches in
Lw(J) and update diagonal sums
G *
C * *
A * *
T * *
C * *
G
G *
C
I
J
For the first 2-word in I,
GC, LGC(J) = {6}.
We can then update
the sum of diagonal
l = i – j = 1 – 6 = -5 to
S-5 := S-5 + 1 = 0 + 1 = 1

244
G *
C * *
A * *
T * *
C * *
G
G *
C
I
J
Next 2-word in I is CA,
for which LCA(J) = {2, 8}.
Two diagonal sums are
updated:
l = i – j = 2 – 2 = 0
S0 := S0 + 1 = 0 + 1 = 1
I = i – j = 2 – 8 = -6
S-6 := S-6 + 1 = 0 + 1 = 1

245
G *
C * *
A * *
T * *
C * *
G
G *
C
I
J
Next 2-word in I is AT,
for which LAT(J) = {3, 9}.
Two diagonal sums are
updated:
l = i – j = 3 – 3 = 0
S0 := S0 + 1 = 1 + 1 = 2
I = i – j = 3 – 9 = -6
S-6 := S-6 + 1 = 1 + 1 = 2

246
After going through the k-words of I, the result is:
l -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Sl 0 0 0 0 4 1 0 0 0 0 4 1 0 0 0 0 0
G *
C * *
A * *
T * *
C * *
G
G *
C
I
J

247
Algorithm for computing diagonal sum of scores
Sl := 0 for all 1 – m l n – 1
Compute Lw(J) for all words w
for i := 1 to n – k – 1 do
w := IiIi+1…Ii+k-1
for j Lw(J) do
l := i – j
Sl := Sl + 1
end
end
Match score is here 1

248
FASTA outline
score matrix
gaps
alignments

249
Rescoring initial regions
p Each high-scoring diagonal chosen in the
previous step is rescored according to a score
matrix
p This is done to find subregions with identities
shorter than k
p Non-matching ends of the diagonal are trimmed
I: C C A T C G C C A T C G
J: C C A A C G C A A T C A
I’: C C A T C G C C A T C G
J’: A C A T C A A A T A A A
75% identity, no 4-word identities
33% identity, one 4-word identity

250
Joining diagonals
p Two offset diagonals can be joined with a gap, if
the resulting alignment has a higher score
p Separate gap open and extension are used
p Find the best-scoring combination of diagonals
High-scoring
diagonals
Two diagonals
joined by a gap

251
FASTA outline
score matrix
gaps
alignments

252
Local alignment in the highest-scoring
region
p Last step of FASTA: perform local
alignment using dynamic
programming around the highest-
scoring
p Region to be aligned covers –w and
+w offset diagonal to the highest-
scoring diagonals
p With long sequences, this region is
typically very small compared to the
whole n x m matrix w
w
Dynamic programming matrix
M filled only for the green region

253
Properties of FASTA
p Fast compared to local alignment using dynamic
programming only
n Only a narrow region of the full matrix is aligned
p Increasing parameter k decreases the number of
hits:
n Increases specificity
n Decreases sensitivity
n Decreases running time
p FASTA can be very specific when identifying long
regions of low similarity
n Specific method does not find many incorrect results
n Sensitive method finds many of the correct results

254
Properties of FASTA
p FASTA looks for initial exact matches to
query sequence
n Two proteins can have very different amino
acid sequences and still be biologically similar
n This may lead into a lack of sensitivity with
diverged sequences

255
Demonstration of FASTA at EBI
p http://www.ebi.ac.uk/fasta/
p Note that parameter ktup in the software
corresponds to parameter k in lectures

256
Note on sequences and alignment
matrices in exercises
p Example solutions to alignment problems
will have sequences arranged like this:
n Perform global alignment of the sequences
p s = AGCTGCGTACT
p t = ATGAGCGTTA
AGCTGCGTACT
ATGAGCGTTA
So if you want to be able to
compare your solution easily
against the example, use this
convention.

257
BLAST
p Search strategies
p FASTA
p BLAST

258
BLAST: Basic Local Alignment Search
Tool
p BLAST (Altschul et al., 1990) and its variants are
some of the most common sequence search tools
in use
p Roughly, the basic BLAST has three parts:
n 1. Find segment pairs between the query sequence and
a database sequence above score threshold (”seed hits”)
n 2. Extend seed hits into locally maximal segment pairs
n 3. Calculate p-values and a rank ordering of the local
alignments
p Gapped BLAST introduced in 1997 allows for gaps
in alignments

259
Finding seed hits
p First, we generate a set of neighborhood
sequences for given k, match score matrix and
threshold T
p Neighborhood sequences of a k-word w include
all strings of length k that, when aligned against
w, have the alignment score at least T
p For instance, let I = GCATCGGC, J =
CCATCGCCATCG and k = 5, match score be 1,
mismatch score be 0 and T = 4

260
Finding seed hits
p I = GCATCGGC, J = CCATCGCCATCG, k = 5,
match score 1, mismatch score 0, T = 4
p This allows for one mismatch in each k-word
p The neighborhood of the first k-word of I, GCATC,
is GCATC and the 15 sequences
A A C A A
CCATC,G GATC,GC GTC,GCA CC,GCAT G
T T T G T

261
Finding seed hits
p I = GCATCGGC has 4 k-words and thus 4x16 =
64 5-word patterns to locate in J
n Occurences of patterns in J are called seed hits
p Patterns can be found using exact search in time
proportional to the sum of pattern lengths +
length of J + number of matches (Aho-Corasick
algorithm)
n Methods for pattern matching are developed on course
58093 String processing algorithms
p Compare this approach to FASTA

262
Extending seed hits: original BLAST
p Initial seed hits are extended into
locally maximal segment pairs
or High-scoring Segment Pairs
(HSP)
p Extensions do not add gaps to the
alignment
p Sequence is extended until the
alignment score drops below the
maximum attained score minus a
threshold parameter value
p All statistically significant HSPs
reported
AACCGTTCATTA
| || || ||
TAGCGATCTTTT
Initial seed hit
Extension
Altschul, S.F., Gish, W., Miller, W., Myers, E. W. and
Lipman, D. J., J. Mol. Biol., 215, 403-410, 1990

263
Extending seed hits: gapped BLAST
p In a later version of BLAST, two
seed hits have to be found on the
same diagonal
n Hits have to be non-overlapping
n If the hits are closer than A
(additional parameter), then they
are joined into a HSP
p Threshold value T is lowered to
achieve comparable sensitivity
p If the resulting HSP achieves a
score at least Sg, a gapped
extension is triggered
Altschul SF, Madden TL, Schäffer AA, Zhang J, Zhang Z, Miller W, and
Lipman DJ, Nucleic Acids Res. 1;25(17), 3389-402, 1997

264
Gapped extensions of HSPs
p Local alignment is performed
starting from the HSP
p Dynamic programming matrix
filled in ”forward” and
”backward” directions (see
figure)
p Skip cells where value would
be Xg below the best
alignment score found so far
Region potentially searched
by the alignment algorithm
HSP
Region searched with score
above cutoff parameter

265
Estimating the significance of results
p In general, we have a score S(D, X) = s for a
sequence X found in database D
p BLAST rank-orders the sequences found by p-
values
p The p-value for this hit is P(S(D, Y) s) where Y
is a random sequence
n Measures the amount of ”surprise” of finding sequence X
p A smaller p-value indicates more significant hit
n A p-value of 0.1 means that one-tenth of random
sequences would have as large score as our result

266
p In BLAST, p-values are computed roughly as
follows
p There are nm places to begin an optimal
alignment in the n x m alignment matrix
p Optimal alignment is preceded by a mismatch
and has t matching (identical) letters
n (Assume match score 1 and mismatch/indel score - )
p Let p = P(two random letters are equal)
p The probability of having a mismatch and then t
matches is (1-p)pt

267
p We model this event by a Poisson distribution
(why?) with mean = nm(1-p)pt
p P(there is local alignment t or longer)
1 – P(no such event)
= 1 – e- = 1 – exp(-nm(1-p)pt)
p An equation of the same form is used in Blast:
p E-value = P(S(D, Y) s) 1 – exp(-nm t) where
> 0 and 0 < < 1
p Parameters and are estimated from data

268
Scoring amino acid
alignments
p We need a way to compute the
score S(D, X) for aligning the
sequence X against database D
p Scoring DNA alignments was
discussed previously
p Constructing a scoring model for
amino acids is more challenging
n 20 different amino acids vs. 4
bases
p Figure shows the molecular
structures of the 20 amino acids
http://en.wikipedia.org/wiki/List_of_standard_amino_acids

269
Scoring amino acid
alignments
p Substitutions between chemically
similar amino acids are more
frequent than between dissimilar
amino acids
p We can check our scoring model
against this
http://en.wikipedia.org/wiki/List_of_standard_amino_acids

270
Score matrices
p Scores s = S(D, X) are obtained from score
matrices
p Let A = A1a2…an and B = b1b2…bn be sequences
of equal length (no gaps allowed to simplify
things)
p To obtain a score for alignment of A and B, where
ai is aligned against bi, we take the ratio of two
probabilities
n The probability of having A and B where the characters
match (match model M)
n The probability that A and B were chosen randomly
(random model R)

271
Score matrices: random model
p Under the random model, the probability
of having X and Y is
where qxi is the probability of occurence of
amino acid type xi
p Position where an amino acid occurs does
not affect its type

272
Score matrices: match model
p Let pab be the probability of having amino acids
of type a and b aligned against each other given
they have evolved from the same ancestor c
p The probability is

273
Score matrices: log-odds ratio score
p We obtain the score S by taking the ratio
of these two probabilities
and taking a logarithm of the ratio

274
Score matrices: log-odds ratio score
p The score S is obtained by summing over
character pair-specific scores:
p The probabilities qa and pab are extracted
from data

275
Calculating score matrices for amino
acids
p Probabilities qa are in
principle easy to obtain:
n Count relative frequencies of
every amino acid in a sequence
database

276
p To calculate pab we can use a
known pool of aligned sequences
p BLOCKS is a database of highly
conserved regions for proteins
p It lists multiply aligned, ungapped
and conserved protein segments
p Example from BLOCKS shows
genes related to human gene
associated with DNA-repair
defect xeroderma pigmentosum
Calculating score matrices for amino
acids
Block PR00851A
ID XRODRMPGMNTB; BLOCK
AC PR00851A; distance from previous block=(52,131)
DE Xeroderma pigmentosum group B protein signature
BL adapted; width=21; seqs=8; 99.5%=985; strength=1287
XPB_HUMAN|P19447 ( 74) RPLWVAPDGHIFLEAFSPVYK 54
XPB_MOUSE|P49135 ( 74) RPLWVAPDGHIFLEAFSPVYK 54
P91579 ( 80) RPLYLAPDGHIFLESFSPVYK 67
XPB_DROME|Q02870 ( 84) RPLWVAPNGHVFLESFSPVYK 79
RA25_YEAST|Q00578 ( 131) PLWISPSDGRIILESFSPLAE 100
Q38861 ( 52) RPLWACADGRIFLETFSPLYK 71
O13768 ( 90) PLWINPIDGRIILEAFSPLAE 100
O00835 ( 79) RPIWVCPDGHIFLETFSAIYK 86
http://blocks.fhcrc.org

277
BLOSUM matrix
p BLOSUM is a score matrix
for amino acid sequences
derived from BLOCKS data
p First, count pairwise
matches fx,y for every amino
acid type pair (x, y)
p For example, for column 3
and amino acids L and W,
we find 8 pairwise matches:
fL,W = fW,L = 8
RPLWVAPD
RPLWVAPR
RPLWVAPN
PLWISPSD
RPLWACAD
PLWINPID
RPIWVCPD

278
p Probability pab is obtained by
dividing fab with the total
number of pairs (note
difference with course book):
p We get probabilities qa by
RPLWVAPD
RPLWVAPR
RPLWVAPN
PLWISPSD
RPLWACAD
PLWINPID
RPIWVCPD
Creating a BLOSUM matrix

279
Creating a BLOSUM matrix
p The probabilities pab and qa can now be plugged
into
to get a 20 x 20 matrix of scores s(a, b).
p Next slide presents the BLOSUM62 matrix
n Values scaled by factor of 2 and rounded to integers
n Additional step required to take into account expected
evolutionary distance
n Described in Deonier’s book in more detail

280
BLOSUM62
A R N D C Q E G H I L K M F P S T W Y V B Z X *
A 4 -1 -2 -2 0 -1 -1 0 -2 -1 -1 -1 -1 -2 -1 1 0 -3 -2 0 -2 -1 0 -4
R -1 5 0 -2 -3 1 0 -2 0 -3 -2 2 -1 -3 -2 -1 -1 -3 -2 -3 -1 0 -1 -4
N -2 0 6 1 -3 0 0 0 1 -3 -3 0 -2 -3 -2 1 0 -4 -2 -3 3 0 -1 -4
D -2 -2 1 6 -3 0 2 -1 -1 -3 -4 -1 -3 -3 -1 0 -1 -4 -3 -3 4 1 -1 -4
C 0 -3 -3 -3 9 -3 -4 -3 -3 -1 -1 -3 -1 -2 -3 -1 -1 -2 -2 -1 -3 -3 -2 -4
Q -1 1 0 0 -3 5 2 -2 0 -3 -2 1 0 -3 -1 0 -1 -2 -1 -2 0 3 -1 -4
E -1 0 0 2 -4 2 5 -2 0 -3 -3 1 -2 -3 -1 0 -1 -3 -2 -2 1 4 -1 -4
G 0 -2 0 -1 -3 -2 -2 6 -2 -4 -4 -2 -3 -3 -2 0 -2 -2 -3 -3 -1 -2 -1 -4
H -2 0 1 -1 -3 0 0 -2 8 -3 -3 -1 -2 -1 -2 -1 -2 -2 2 -3 0 0 -1 -4
I -1 -3 -3 -3 -1 -3 -3 -4 -3 4 2 -3 1 0 -3 -2 -1 -3 -1 3 -3 -3 -1 -4
L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4 -2 2 0 -3 -2 -1 -2 -1 1 -4 -3 -1 -4
K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5 -1 -3 -1 0 -1 -3 -2 -2 0 1 -1 -4
M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5 0 -2 -1 -1 -1 -1 1 -3 -1 -1 -4
F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6 -4 -2 -2 1 3 -1 -3 -3 -1 -4
P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7 -1 -1 -4 -3 -2 -2 -1 -2 -4
S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4 1 -3 -2 -2 0 0 0 -4
T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5 -2 -2 0 -1 -1 0 -4
W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11 2 -3 -4 -3 -2 -4
Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7 -1 -3 -2 -1 -4
V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4 -3 -2 -1 -4
B -2 -1 3 4 -3 0 1 -1 0 -3 -4 0 -3 -3 -2 0 -1 -4 -3 -3 4 1 -1 -4
Z -1 0 0 1 -3 3 4 -2 0 -3 -3 1 -1 -3 -1 0 -1 -3 -2 -2 1 4 -1 -4
X 0 -1 -1 -1 -2 -1 -1 -1 -1 -1 -1 -1 -1 -1 -2 0 0 -2 -1 -1 -1 -1 -1 -4
* -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 -4 1

281
Using BLOSUM62 matrix
MQLEANADTSV
| | |
LQEQAEAQGEM
= 2 + 5 – 3 – 4 + 4 + 0 + 4 + 0 – 2 + 0 +
1
= 7

282
Demonstration of BLAST at NCBI
p http://www.ncbi.nlm.nih.gov/BLAST/

Introduction to
Bioinformatics
Lecture 4:
Genome rearrangements

284
Why study genome rearrangements?
p Provide insight into evolution of species
p Fun algorithmic problem!
p Structure of this lecture:
n The biological phenomenon
n How to computationally model it?
n How to compute interesting things?
n Studying the phenomenon using existing tools
(continued in exercises)

285
Genome rearrangements as an
algorithmic problem

286
Background
p Genome sequencing enables us to
compare genomes of two or more
different species
n -> Comparative genomics
p Basic observation:
n Closely related species (such as human and
mouse) can be almost identical in terms of
genome contents...
n ...but the order of genomic segments can be
very different between species

287
Synteny blocks and segments
p Synteny – derived from Greek ’on the
same ribbon’ – means genomic segments
located on the same chromosome
n Genes, markers (any sequence)
p Synteny block (or syntenic block)
n A set of genes or markers that co-occur
together in two species
p Synteny segment (or syntenic segment)
n Syntenic block where the order of genes or
markers is preserved

288
Synteny blocks and segments
Chromosome i, species B
Chromosome j, species C
Synteny segment
Synteny block
Homologs
of the same
gene

289
Observations from sequencing
1. Large chromosome inversions and
translocations (we’ll get to these shortly)
are common
n ...Even between closely related species
2. Chromosome inversions are usually
symmetric around the origin of DNA
replication
3. Inversions are less common within
species...

290
What causes rearrangements?
p RecA, Recombinase A,
is a protein used to
repair chromosomal
damage
p It uses a duplicate
copy of the damaged
sequence as template
p Template is usually a
homologous sequence
on a sister
chromosome
Diarmaid Hughes: Evaluating genome dynamics: the constraints on
rearrangements within bacterial genomes, Genome Biology 2000, 1

291
Chromosomes: recap
p Linear chromosomes
n Eukaryotes (mostly)
p Circular chromosomes
n Prokaryotes (mostly)
n Mitochondria
chromatid
centromere
gene 1
gene 3
gene 2
Also double-stranded: genes can be
found on both strands (orientations)

292
What effects does RecA have on
genome?
p Repeated sequences cause RecA to fail to
choose correct recombination start
position
p This leads to
n Tandem duplications
n Translocations
n Inversions
Repeat 1 Repeat 2
RecA
?
Damaged sequence

293
X, Y, Z and W are repeats of
the same sequence.
a, b, c and d are sequences on genome
bounded by repeats.
In a tandem duplication example,
RecA recombines a sequence that
starts from Y instead of Z after Z.
This leads to duplication of segment Y-Z.

294
Recombination of two
repeat sequences in the
same chromosome can
lead to a fragment translocation
Here sequence d is translocated

295
Inversion happens when two
sequences of opposite orientations
are recombined.

296
Example: human vs mouse genome
p Human and mouse genomes share
thousands of homologous genes, but they
are
n Arranged in different order
n Located in different chromosomes
p Examples
n Human chromosome 6 contains elements from
six different mouse chromosomes
n Analysis of X chromosome indicates that
rearrangements have happened primarily
within chromosome

299
Representing genome rearrangments
p When comparing two genomes, we can
find homologous sequences in both using
BLAST, for example
p This gives us a map between sequences in
both genomes

300
Representing genome rearrangments
p We assign numbers 1,...,n to
the found homologous
sequences
p By convention, we number the
sequences in the first genome
by their order of appearance
in chromosomes
p If the homolog of i is in
reverse orientation, it receives
number –i (signed data)
p For example, consider human
vs mouse gene numbering on
the right
(il10)
9
(at3)
-6
(pdc)
8
(lamc1)
-7
(lamc1)
7
(pdc)
-8
(at3)
6
(il10)
-9
(pklr)
5
(pax3)
15
(gba)
4
(fn1)
14
(ngfb)
3
(cd28)
13
(nras)
2
(inpp1)
12
(gnat2)
1
Mouse
Human
List order corresponds to
physical order on chromosomes!

301
Permutations
p The basic data structure in the study of
genome rearrangements is permutation
p A permutation of a sequence of n numbers
is a reordering of the sequence
p For example, 4 1 3 2 5 is a permutation of
1 2 3 4 5

302
Genome rearrangement problem
p Given two genomes (set of markers), how
many
n duplications,
n inversions and
n translocations
do we need to do to transform the first
genome to the second?
Minimum number of operations?
What operations? Which order?

303
6 1 2 3 4 5 1 2 3 4 5 6
#duplications?
#inversions?
#translocations?

304
6 1 2 3 4 5 1 2 3 4 5 6
1 2 3 4 5 6
Keep in mind, that the two genomes
have been evolved from a common
ancestor genome!
Permutation
of 1,...,6

305
Genome rearrangements using reversals
(=inversions) only
p Lets consider a simpler problem where we just
study reversals with unsigned data
p A reversal p(i, j) reverses the order of the
segment i i+1 ... j-1 j (indexing starts from 1)
p For example, given permutation
6 1 2 3 4 5 and reversal p(3, 5) we get
permutation 6 1 4 3 2 5
...note that we do not care about exact positions on the genome
p(3, 5)

306
Reversal distance problem
p Find the shortest series of reversals that, given
a permutation , transforms it to the identity
permutation (1, 2, ..., n)
p This quantity is denoted by d( )
p Reversal distance for a pair of chromosomes:
n Find synteny blocks in both
n Number blocks in the first chromosome to identity
n Set to correspond matching of second chromosome’s
blocks against the first
n Find reversal distance

307
Reversal distance problem: discussion
p If we can find the minimal series of
reversals for some pair of genomes
n Is that what happened during evolution?
n If not, is it the correct number of reversals?
p In any case, reversal distance gives us a
measure of evolutionary distance between
the two genomes and species

308
Solving the problem by sorting
p Our first approach to solve the reversal
distance problem:
n Examine each position i of the permutation
n At each position, if i i, do a reversal such
that i = i
p This is a greedy approach: we try to
choose the best option at each step

309
Simple reversal sort: example
6 1 2 3 4 5 -> 1 6 2 3 4 5 -> 1 2 6 3 4 5 -> 1 2 3 4 6 5
-> 1 2 3 4 5 6
Reversal series: p(1,2), p(2,3), p(3,4), p(5,6)
Is d(6 1 2 3 4 5) then 4?
6 1 2 3 4 5 -> 5 4 3 2 1 6 -> 1 2 3 4 5 6
D(6 1 2 3 4 5) = 2

310
Pancake flipping problem
p No pancake made by
the chef is of the
same size
p Pancakes need to be
rearranged before
delivery
p Flipping operation:
take some from the
top and flip them over
p This corresponds to
always reversing the
sequence prefix
1 2 3 6 4 5 -> 6 3 2 1 4 5 ->
5 4 1 2 3 6 -> 3 2 1 4 5 6 ->
1 2 3 4 5 6

311
How good is simple reversal sort?
p Not so good actually
p It has to do at most n-1 reversals with
permutation of length n
p The algorithm can return a distance that is
as large as (n – 1)/2 times the correct
result d( )
n For example, if n = 1001, result can be as bad
as 500 x d( )

312
Estimating reversal distance by cycle
decomposition
p We can estimate d( ) by cycle
decomposition
p Lets represent permutation = 1 2 4 5 3
with the following graph
where edges correspond to adjacencies
(identity, permutation F)
1 2 4 5 3
0 6

313
Estimating reversal distance by cycle
decomposition
p Cycle decomposition: a set of cycles that
n have edges with alternating colors
n do not share edges with other cycles (=cycles
are edge disjoint)
1 2 4 5 3
0 6
1 2 4 5

314
Cycle decompositions
p Let c( ) the maximum number of alternating,
edge-disjoint cycles in the graph representation
of
p The following formula allows estimation of d( )
n d( ) n + 1 – c( ), where n is the permutation length
1 2 4 5 3
0 6
1 2 4 5
d( ) 5 + 1 – 4 = 2
Claim in Deonier: equality holds for ”most of the usual and
interesting biological systems.

315
Cycle decompositions
p Cycle decomposition is NP-complete
n We cannot solve the general problem exactly
for large instances
p However, with signed data the problem
becomes easy
n Before going into signed data, lets discuss
another algorithm for the general case

316
Computing reversals with breakpoints
p Lets investigate a better way to compute
reversal distance
p First, some concepts related to
permutation 1 2,,, n-1 n
n Breakpoint: two elements i and i+1 are a
breakpoint, if they are not consecutive
numbers
n Adjacency: if i and i+1 are consecutive, they
are called adjacency

317
Breakpoints and adjacencies
2 1 3 4 5 8 7 6
This permutation contains
four breakpoints begin-2, 13, 58, 6-end and
five adjacencies 21, 34, 45, 87, 76
Breakpoints

318
Breakpoints
p Each breakpoint in permutation needs to be
removed to get to the identity permutation (=our
target)
n Identity permutation does not contain any breakpoints
p First and last positions special cases
p Note that each reversal can remove at most two
breakpoints
p Denote the number of breakpoints by b( )
2 1 3 4 5 8 7 6 b( ) = 4

319
Breakpoint reversal sort
p Idea: try to remove as many breakpoints
as possible (max 2) in every step
1. While b( ) > 0
2. Choose reversal p that removes most breakpoints
3. Perform reversal p to
4. Output
5. return

320
Breakpoint removal: example
8 2 7 6 5 1 4 3 b( ) = 6
2 8 7 6 5 1 4 3 b( ) = 5
2 3 4 1 5 6 7 8 b( ) = 3
4 3 2 1 5 6 7 8 b( ) = 2
1 2 3 4 5 6 7 8 b( ) = 0

321
Breakpoint removal
p The previous algorithm needs refinement
to be correct
p Consider the following permutation:
1 5 6 7 2 3 4 8
p There is no reversal that decreases the
number of breakpoints!
p See Jones & Pevzner for detailed
description on this

322
Breakpoint removal
p Reversal can only decrease breakpoint
count if permutation contains decreasing
strips
1 5 6 7 2 3 4 8
1 5 6 7 4 3 2 8
1 2 3 4 7 6 5 8
Increasing strip
Decreasing strip
Strip: maximal segment without breakpoints

323
Improved breakpoint reversal sort
1. While b( ) > 0
2. If has a decreasing strip
3. Do reversal p that removes most BPs
4. Else
5. Reverse an increasing strip
6. Output
7. return

324
Is Improved BP removal enough?
p The algorithm works pretty well:
n It produces a result that is at most four times
worse than the optimal result
n ...is this good?
p We considered only reversals
p What about translocations & duplications?

325
Translocations via reversals
1 2 3 4 5 6 7 8
1 5 6 7 8 2 3 4
1 4 3 2 8 7 6 5
1 2 3 4 8 7 6 5
1 2 3 4 5 6 7 8
Translocation of 2,3,4
p(2,8)
p(2,4)
p(5,8)

326
Genome rearrangements with reversals
p With unsigned data, the problem of finding
minimum reversal distances is NP-
complete
n Why is this so if sorting is easy?
p An algorithm has been developed that
achieves 1.375-approximation
p However, reversal distance in signed data
can be computed quickly!
n It takes linear time w.r.t. the length of
permutation (Bader, Moret, Yan, 2001)

327
Cycle decomposition with signed data
p Consider the following two permutations
that include orientation of markers
n J: +1 +5 -2 +3 +4
n K: +1 -3 +2 +4 -5
p We modify this representation a bit to
include both endpoints of each marker:
n J’: 0 1a 1b 5a 5b 2b 2a 3a 3b 4a 4b 6
n K’: 0 1a 1b 3b 3a 2a 2b 4a 4b 5b 5a 6

328
Graph representation of J’ and K’
p Drawn online in lecture!

329
Multiple chromosomes
p In unichromosomal genomes, inversion
(reversal) is the most common operation
p In multichromosomal genomes,
inversions, translocations, fissions and
fusions are most common

330
p Lets represent multichromosomal genome
as a set of permutations, with $ denoting
the boundary of a chromosome:
5 9 $
1 3 2 8 $
7 6 4 $
This notation is frequently used in software
used to analyse genome rearrangements.
Chr 1
Chr 2
Chr 3

331
p Note that when dealing with multiple
chromosomes, you need to specify
numbering for elements on both genomes

332
Reversals & translocations
p Reversal p( , i, j)
p Translocation p( , , i, j)
i
j
Translocation

333
Fusions & fissions
p Fusion: merging of two chromosomes
p Fission: chromosome is split into two
chromosomes
p Both events can be represented with a
translocation

334
Fusion
p Fusion by translocation p( , , n+1, 1)
i = n + 1
j = 1
Fusion

335
Fission
p Fission by translocation p( , , i, 1)
i
Empty chromosome
Fission

336
Algorithms for general genomic distance
problem
p Hannenhalli, Pevzner: Transforming Men into
Mice (polynomial algorithm for genomic distance
problem), 36th Annual IEEE Symposium on
Foundations of Computer Science, 1995

337
Human & mouse revisited
p Human and mouse are separated by about
75-83 million years of evolutionary history
p Only a few hundred rearrangements have
happened after speciation from the
common ancestory
p Pevzner & Tesler identified in 2003 for 281
synteny blocks a rearrangement from
mouse to human with
n 149 inversions
n 93 translocations
n 9 fissions

338
Discussion
p Genome rearrangement events are very
rare compared to, e.g., point mutations
n We can study rearrangement events further
back in the evolutionary history
p Rearrangements are easier to detect in
comparison to many other genomic events
p We cannot detect homologs 100%
correctly so the input permutation can
contain errors

339
Discussion
p Genome rearrangement is to some degree
constrained by the number and size of
repeats in a genome
n Notice how the importance of genomic repeats
pops up once again
p Sequencing gives us (usually) signed data
so we can utilize faster algorithms
p What if there are more than one optimal
solution?

340
Two different genome rearrangement scenarios
giving the same result.

341
GRIMM demonstration
Glenn Tesler, GRIMM: genome rearrangements web server.
Bioinformatics, 2002,

342
GRIMM file format
# useful comment about first genome
# another useful comment about it
>name of first genome
1 -4 2 $ # chromosome 1
-3 5 6 # chromosome 2
>name of second genome
5 -3 $
6 $
2 -4 1 $
http://grimm.ucsd.edu/GRIMM/grimm_instr.html
GRIMM supports analysis of
one, two or more genomes

Introduction to
Bioinformatics
Phylogenetic trees

344
Inferring the Past: Phylogenetic
Trees
p Parsimony and distance methods
p Models for mutations and estimation of
distances
p Maximum likelihood methods

345
Phylogeny
p We want to study ancestor-
descendant relationships, or
phylogeny, among groups of
organisms
p Groups are called taxa (singular:
taxon)
p Organisms are usually called
operational taxonomic units or
OTUs in the context of
phylogeny

346
Phylogenetic trees
p Leaves (external nodes)
~ species, observed
(OTUs)
p Internal nodes ~
ancestral
species/divergence
events, not observed
p Unrooted tree does not
specify ancestor-
descendant relationships
beyond the observation
”leaves are not
ancestors”
1
2
3
4
5
6
7
8
Unrooted tree with 5
leaves and 3 internal
nodes.
Is node 7 ancestor of node
6?

347
Phylogenetic trees
p Rooting a tree specifies all
ancestor-descendant
relationships in the tree
p Root is the ancestor to the
other species
p There are n-1 ways to root
a tree with n nodes
1
2
3
4
5
6
7
8
R1 R2
2 3 4 5
1
6
7
8
R1
2 3 4
5
1
6
7
8
R2
r
o
o
t
(
R
1
)
r
o
o
t
(
R
2
)

348
Questions
p Can we enumerate all possible
phylogenetic trees for n species (or
sequences?)
p How to score a phylogenetic tree with
respect to data?
p How to find the best phylogenetic tree
given data?

349
Finding the best phylogenetic tree:
naive method
p How can we find the phylogenetic tree
that best represents the data?
p Naive method: enumerate all possible
trees
p How many different trees are there of n
species?
p Denote this number by bn

350
Enumerating unordered trees
p Start with the only
unordered tree with 3
leaves (b3 = 1)
p Consider all ways to add
a leaf node to this tree
p Fourth node can be added
to 3 different branches
(edges), creating 1 new
internal branch
p Total number of branches
is n external and n – 3
internal branches
p Unrooted tree with n
leaves has 2n – 3 branches
1 2
3
1 2
3
4
1 2
3
4
1 2
3
4

351
Enumerating unordered trees
p Thus, we get the number of unrooted trees
bn = (2(n – 1) – 3)bn-1 = (2n – 5)bn-1
= (2n – 5) * (2n – 7) * …* 3 * 1
= (2n – 5)! / ((n-3)!2n-3), n > 2
p Number of rooted trees b’n is
b’n = (2n – 3)bn = (2n – 3)! / ((n-2)!2n-2),
n > 2
that is, the number of unrooted trees times the
number of branches in the trees

352
Number of possible rooted and
unrooted trees
8.20E+021
2.22E+020
20
4.95E+038
8.69E+036
30
34459425
2027025
10
2027025
135135
9
135135
10395
8
10395
954
7
945
105
6
105
15
5
15
3
4
3
1
3
b’n
Bn
n

353
Too many trees?
p We can’t construct and evaluate every
phylogenetic tree even for a smallish
number of species
p Better alternative is to
n Devise a way to evaluate an individual tree
against the data
n Guide the search using the evaluation criteria
to reduce the search space

354
Trees (chapter 12)
distances

355
Parsimony method
p The parsimony method finds the tree that
explains the observed sequences with a
minimal number of substitutions
p Method has two steps
n Compute smallest number of substitutions for
a given tree with a parsimony algorithm
n Search for the tree with the minimal number of
substitutions

356
Parsimony: an example
p Consider the following short sequences
1 ACTTT
2 ACATT
3 AACGT
4 AATGT
5 AATTT
p There are 105 possible rooted trees for 5
sequences
p Example: which of the following trees
explains the sequences with least number
of substitutions?

357
3
AACGT
4
AATGT
5
AATTT
2
ACATT
1
ACTTT
6 AATGT
7 AATTT
8 ACTTT
9 AATTT
T->C
T->G
T->A
A->C
This tree explains the sequences
with 4 substitutions

358
3
AACGT
4
AATGT
5
AATTT
2
ACATT
1
ACTTT
6 AATGT
7 AATTT
8 ACTTT
9 AATTT
T->C
T->G
T->A
A->C
3
AACGT
4
AATGT
5
AATTT
2
ACATT
1
ACTTT
6 ACCTT
C->T
7 AACGT
8 AATGT
9 AATTT
G->T
T->C
T->G
A->C
C->A
6
substitutions…
First tree is
more
parsimonious!
4
substitutions…

359
Computing parsimony
p Parsimony treats each site (position in a
sequence) independently
p Total parsimony cost is the sum of parsimony
costs (=required substitutions) of each site
p We can compute the minimal parsimony cost for
a given tree by
n First finding out possible assignments at each node,
starting from leaves and proceeding towards the root
n Then, starting from the root, assign a letter at each
node, proceeding towards leaves

360
Labelling tree nodes
p An unrooted tree with n leaves contains 2n-1
nodes altogether
p Assign the following labels to nodes in a rooted
tree
n leaf nodes: 1, 2, …, n
n internal nodes: n+1, n+2, …, 2n-1
n root node: 2n-1
p The label of a child node is always
smaller than the label of the
parent node
2 3 4 5
1
6
8
7
9

361
Parsimony algorithm: first phase
p Find out possible assignments at every node for each
site u independently. Denote site u in sequence i by
si,u.
For i := 1, …, n do
Fi := {si,u} % possible assignments at node i
Li := 0 % number of substitutions up to node i
For i := n+1, …, 2n-1 do
Let j and k be the children of node i
If Fj Fk =
then Li := Lj + Lk + 1, Fi := Fj Fk
else Li := Lj + Lk, Fi := Fj Fk

362
3
AACGT
4
AATGT
5
AATTT
2
ACATT
1
ACTTT
Choose u = 3 (for example, in general we do this for all sites)
F1 := {T}
L1 := 0
F2 := {A}
L2 := 0
F3 := {C}, L3 := 0
F4 := {T}, L4 := 0
F5 := {T}, L5 := 0
6
7
8
9

363
3
AACGT
4
AATGT
5
AATTT
2
ACATT
1
ACTTT
6 {C,T}
7 T
8 {A, T}
9 T
F8 := F1 F2 = {A, T}
L8 := L1 + L2 + 1 = 1
F6 := F3 F4 = {C, T}
L6 := L3 + L4 + 1 = 1
F7 := F5 F6 = {T}
L7 := L5 + L6 = 1
F9 := F7 F8 = {T}
L9 := L7 + L8 = 2
Parsimony cost for site 3 is 2

364
Parsimony algorithm: second phase
p Backtrack from the root and assign x Fi
at each node
p If we assigned y at parent of node i and y
Fi, then assign y
p Else assign x Fi by random

365
Parsimony algorithm: second phase
3
AACGT
4
AATGT
5
AATTT
2
ACATT
1
ACTTT
6 {C,T}
7 T
8 {A,T}
9 T
At node 6, the
algorithm assigns T
because T was
assigned to parent
node 7 and T F6.
T is assigned to node 8
for the same reason.
The other nodes have
only one possible letter
to assign

366
Parsimony algorithm
3
AACGT
4
AATGT
5
AATTT
2
ACATT
1
ACTTT
6 T
7 T
8 T
9 T
First and second phase are
repeated for each site in
the sequences,
summing the parsimony
costs at each site

367
Properties of parsimony algorithm
p Parsimony algorithm requires that the sequences
are of same length
n First align the sequences against each other and,
optionally, remove indels
n Then compute parsimony for the resulting sequences
n Indels (if present) considered as characters
p Is the most parsimonious tree the correct tree?
n Not necessarily but it explains the sequences with least
number of substitutions
n We can assume that the probability of having fewer
mutations is higher than having many mutations

368
Finding the most parsimonious tree
p Parsimony algorithm calculates the
parsimony cost for a given tree…
p …but we still have the problem of finding
the tree with the lowest cost
p Exhaustive search (enumerating all trees)
is in general impossible
p More efficient methods exist, for example
n Probabilistic search
n Branch and bound

369
Branch and bound in parsimony
p We can exploit the fact that adding edges
to a tree can only increase the parsimony
cost
1
AATGT
2
AATTT
3
AACGT
1
AATGT
2
AATTT
{T}
{T}
{C, T}
cost 0 cost 1

370
Branch and bound in parsimony
Branch and bound is
a general search
strategy where
p Each solution is
potentially generated
p Track is kept of the
best solution found
p If a partial solution
cannot achieve better
score, we abandon
the current search
path
In parsimony…
p Start from a tree with 1
sequence
p Add a sequence to the tree
and calculate parsimony
cost
p If the tree is complete,
check if found the best tree
so far
p If tree is not complete and
cost exceeds best tree
cost, do not continue
adding edges to this tree

371
Branch and bound example
1 1 2 3
1 2 3
1 2 4
…
Complete tree:
compute parsimony
cost
Example with 4 sequences
Partial tree:
Compute parsimony cost
and compare against best
so far;
Do not continue expansion
if above cost of the best tree
3
1 2 4
2
1 3 4
2
1 3 4
1 3
2
1 3

372
Distance methods
p The parsimony method works on sequence
(character string) data
p We can also build phylogenetic trees in a
more general setting
p Distance methods work on a set of
pairwise distances dij for the data
p Distances can be obtained from
phenotypes as well as from genotypes
(sequences)

373
Distances in a phylogenetic tree
p Distance matrix D = (dij)
gives pairwise distances
for leaves of the
phylogenetic tree
p In addition, the
phylogenetic tree will
now specify distances
between leaves and
internal nodes
n Denote these with dij as
well
2 3 4 5
1
6
7
8
Distance dij states how
far apart species i and j
are evolutionary (e.g.,
number of mismatches in
aligned sequences)

374
Distances in evolutionary context
p Distances dij in evolutionary context
satisfy the following conditions
n Symmetry: dij = dji for each i, j
n Distinguishability: dij 0 if and only if i j
n Triangle inequality: dij dik + dkj for each i, j, k
p Distances satisfying these conditions are called
metric
p In addition, evolutionary mechanisms may
impose additional constraints on the distances
additive and ultrametric distances

375
Additive trees
p A tree is called additive, if the distance
between any pair of leaves (i, j) is the
sum of the distances between the leaves
and a node k on the shortest path from i
to j in the tree
dij = dik + djk
p ”Follow the path from the leaf i to the leaf
j to find the exact distance dij between the
leaves.”

376
Additive trees: example
0
2
4
4
D
2
0
4
4
C
4
4
0
2
B
4
4
2
0
A
D
C
B
A
A
B
C
D
1
1
2 1
1

377
Ultrametric trees
p A rooted additive tree is called an ultrametric
tree, if the distances between any two leaves i
and j, and their common ancestor k are equal
dik = djk
p Edge length dij corresponds to the time elapsed
since divergence of i and j from the common
parent
p In other words, edge lengths are measured by a
molecular clock with a constant rate

378
Identifying ultrametric data
p We can identify distances to be ultrametric
by the three-point condition:
D corresponds to an ultrametric tree if
and only if for any three species i, j and
k, the distances satisfy dij max(dik, dkj)
p If we find out that the data is ultrametric, we can
utilise a simple algorithm to find the
corresponding tree

379
Ultrametric trees
9
8
7
5 4 3 2 1
6
Observation time
Time

380
Ultrametric trees
9
8
7
5 4 3 2 1
6
Observation time
Time
Only vertical segments of the
tree have correspondence to
some distance dij:
Horizontal segments act as
connectors.
d8,9

381
Ultrametric trees
9
8
7
5 4 3 2 1
6
Observation time
Time
dik = djk for any two leaves
i, j and any ancestor k of
i and j

382
Ultrametric trees
9
8
7
5 4 3 2 1
6
Observation time
Time
Three-point condition: there are
no leafs i, j for which dij > max(dik, djk)
for some leaf k.

383
UPGMA algorithm
p UPGMA (unweighted pair group method
using arithmetic averages) constructs a
phylogenetic tree via clustering
p The algorithm works by at the same time
n Merging two clusters
n Creating a new node on the tree
p The tree is built from leaves towards the
root
p UPGMA produces a ultrametric tree

384
Cluster distances
p Let distance dij between clusters Ci and Cj
be
that is, the average distance between
points (species) in the cluster.

385
UPGMA algorithm
p Initialisation
n Assign each point i to its own cluster Ci
n Define one leaf for each sequence, and place it at height
zero
p Iteration
n Find clusters i and j for which dij is minimal
n Define new cluster k by Ck = Ci Cj, and define dkl for all l
n Define a node k with children i and j. Place k at height dij/2
n Remove clusters i and j
p Termination:
n When only two clusters i and j remain, place root at height dij/2

390
1 2
3
4
5
1 2 4 5
6 7
8
3
9

391
UPGMA implementation
p In naive implementation, each iteration
takes O(n2) time with n sequences =>
algorithm takes O(n3) time
p The algorithm can be implemented to take
only O(n2) time (see Gronau & Moran,
2006, for a survey)

392
Problem solved?
p We now have a simple algorithm which finds a
ultrametric tree
n If the data is ultrametric, then there is exactly one
ultrametric tree corresponding to the data (we skip the
proof)
n The tree found is then the ”correct” solution to the
phylogeny problem, if the assumptions hold
p Unfortunately, the data is not ultrametric in
practice
n Measurement errors distort distances
n Basic assumption of a molecular clock does not hold
usually very well

393
Incorrect reconstruction of non-
ultrametric data by UPGMA
1
2 3
4
1 2 3
4
Tree which corresponds
to non-ultrametric
distances
Incorrect ultrametric reconstruction
by UPGMA algorithm

394
Checking for additivity
p How can we check if our data is additive?
p Let i, j, k and l be four distinct species
p Compute 3 sums: dij + dkl, dik + djl, dil
+ djk

395
Four-point condition
i
j l
k i
j l
k i
j l
k
dik
djl
dil
djk
dij dkl
p The sums are represented by the three figures
n Left and middle sum cover all edges, right sum does not
p Four-point condition: i, j, k and l satisfy the four-
point condition if two of the sums dij + dkl, dik +
djl, dil + djk are the same, and the third one is
smaller than these two

396
Checking for additivity
p An n x n matrix D is additive if and only if
the four point condition holds for every 4
distinct elements 1 i, j, k, l n

397
Finding an additive phylogenetic tree
p Additive trees can be found with, for example,
the neighbor joining method (Saitou & Nei, 1987)
p The neighbor joining method produces unrooted
trees, which have to be rooted by other means
n A common way to root the tree is to use an outgroup
n Outgroup is a species that is known to be more distantly
related to every other species than they are to each
other
n Root node candidate: position where the outgroup would
join the phylogenetic tree
p However, in real-world data, even additivity
usually does not hold very well

398
Neighbor joining algorithm
p Neighbor joining works in a similar fashion
to UPGMA
n Find clusters C1 and C2 that minimise a
function f(C1, C2)
n Join the two clusters C1 and C2 into a new
cluster C
n Add a node to the tree corresponding to C
n Assign distances to the new branches
p Differences in
n The choice of function f(C1, C2)
n How to assign the distances

399
p Recall that the distance dij for clusters Ci and Cj
was
p Let u(Ci) be the separation of cluster Ci from
other clusters defined by
where n is the number of clusters.

400
p Instead of trying to choose the clusters Ci
and Cj closest to each other, neighbor
joining at the same time
n Minimises the distance between clusters Ci and
Cj and
n Maximises the separation of both Ci and Cj
from other clusters

401
p Initialisation as in UPGMA
p Iteration
n Find clusters i and j for which dij – u(Ci) – u(Cj) is minimal
n Define new cluster k by Ck = Ci Cj, and define dkl for all l
n Define a node k with edges to i and j. Remove clusters i and j
n Assign length ½ dij + ½ (u(Ci) – u(Cj)) to the edge i -> k
n Assign length ½ dij + ½ (u(Cj) – u(Ci)) to the edge j -> k
p Termination:
n When only one cluster remains

402
Neighbor joining algorithm: example
a b c d
a 0 6 7 5
b 0 11 9
c 0 6
d 0
i u(i)
a (6+7+5)/2 = 9
b (6+11+9)/2 = 13
c (7+11+6)/2 = 12
d (5+9+6)/2 = 10
i,j dij – u(Ci) – u(Cj)
a,b 6 - 9 - 13 = -16
a,c 7 - 9 - 12 = -14
a,d 5 - 9 - 10 = -14
b,c 11 - 13 - 12 = -14
b,d 9 - 13 - 10 = -14
c,d 6 - 12 - 10 = -16
Choose either pair
to join

403
Neighbor joining algorithm: example
a b c d
a 0 6 7 5
b 0 11 9
c 0 6
d 0
i u(i)
a (6+7+5)/2 = 9
b (6+11+9)/2 = 13
c (7+11+6)/2 = 12
d (5+9+6)/2 = 10
i,j dij – u(Ci) – u(Cj)
a,b 6 - 9 - 13 = -16
a,c 7 - 9 - 12 = -14
a,d 5 - 9 - 10 = -14
b,c 11 - 13 - 12 = -14
b,d 9 - 13 - 10 = -14
c,d 6 - 12 - 10 = -16
a b c d
e
dae = ½ 6 + ½ (9 – 13) = 1
dbe = ½ 6 + ½ (13 – 9) = 5
dbe
dae
This is the first step only…

404
Trees (chapter 12)
distances
n These parts of the book is skipped on this
course (see slides of 2007 course for material
on these topics)
n No questions in exams on these topics!

405
Problems with tree-building
p Assumptions
n Sites evolve independently of one other
n (Sites evolve according to the same stochastic
model; not really covered this year)
n The tree is rooted
n The sequences are aligned
n Vertical inheritance

406
Additional material on phylogenetic
trees
p Durbin, Eddy, Krogh, Mitchison: Biological
sequence analysis
p Jones, Pevzner: An introduction to
bioinformatics algorithms
p Gusfield: Algorithms on strings, trees, and
sequences
p Course on phylogenetic analyses in Spring
2009

Introduction to
Bioinformatics
Wrap-up

408
Topics
Sequencing
-Sanger
-High-throughput
atgagccaag ttccgaacaa ggattcgcgg
gtcggtaaag agcattggaa cgtcggagat
aagaagcgga tgaatttccc cataacgcca
gtggaagaga aggaggcggg cctcccgatc
actccggccc gaagggttga gagtacccca
gaaatcacct ccagaggacc ccttcagcga
catagcgata ggaggggatg ctaggagttg
Biodatabases
BLAST query
gtggaagagaaggaggcgg…
gaagggttgagagtacccca...
ccagaggaccccttcagcga…
ggaggggatgctaggagttg…
Sequence
assembler
Contigs
=> Genomes
Reads
Similar
sequences = homologs?
Proteomics
Gene expression
analysis
DNA chips
Protein gels
MS/MS techniques
Comparative genomics
- Phylogenetics
- Genome rearrangements
- …
Systems
Biology
2010: Human genome
in 15 minutes!
atgagccaag
aagaagcgga
cagcggaaga

409
Exams
p Course exam Wednesday 15 October
16.00-19.00 Exactum A111
p Separate exams
n Tue 18 November 16.00-20.00 Exactum A111
n Fri 16 January 16.00-20.00 Exactum A111
n Tue 31 March 16.00-20.00 Exactum A111
p Check exam date and place before taking
the exam! (previous week or so)

410
Exam regulations
p If you are late more than 30 min, you
cannot take the exam
p You are not allowed to bring material such
as books or lecture notes to the exam
p Allowed stuff: blank paper (distributed in
the exam), pencils, pens, erasers,
calculators, snacks
p Bring your student card or other id!

411
Grading
p Grading: on the scale 0-5
n To get the lowest passing grade 1, you need to get at
least 30 points out of 60 maximum
p Course exam gives you maximum of 48 points
p Note: if you take the first separate exam, the
best of the following options will be considered:
n Exam gives you max 48 points, exercises max 12 points
n Exam gives you max 60 points
p In second and subsequent separate exams, only
the 60 point option is in use

412
Exercise points
p Max. marks: 31
p 80% of 31 ~= 24 marks -> 12 points
p 2 marks = 1 point

413
Topics covered by exams
p Exams cover everything presented in lectures
(exception: biological background not covered)
p Word distributions and occurrences (course book
chapters 2-3)
p Genome rearrangements (chapter 5)
p Sequence alignment (chapter 6)
p Rapid alignment methods: FASTA and BLAST
(chapter 7)
p Sequencing and sequence assembly (chapter 8)

414
Topics covered by exams
p Similarity, distance and clustering
(chapter 10)
p Expression data analysis (chapter 11)
p Phylogenetic trees (chapter 12)
p Systems biology: modelling biological
networks (no chapter in course book)

415
Bioinformatics courses in 2008
p Biological sequence analysis (II period,
Kumpula)
n Focus on probabilistic methods: Hidden Markov
Models, Profile HMMs, finding regulatory
elements, …
p Modeling of biological networks (20-
24.10., TKK)
n Biochemical network modelling and parameter
estimation in biochemical networks using
mechanistic differential equation models.

416
Bioinformatics courses in autumn 2008
p Bayesian paradigm in genetic
bioinformatics (II period, Kumpula)
n Applications of Bayesian approach in computer
programs and data analysis of
p genetic past,
p phylogenetics,
p coalescence,
p relatedness,
p haplotype structure,
p disease gene associations.

417
Bioinformatics courses in autumn 2008
p Statistical methods in genetics (II period,
Kumpula)
n Introduction to statistical methods in gene
mapping and genetic epidemiology.
n Basic concepts of linkage and association
analysis as well as some concepts of
population genetics will be covered.

418
Bioinformatics courses in Spring 2009
p Practical Course in Biodatabases (III
period, Kumpula)
p High-throughput bioinformatics (III-IV
periods, TKK)
p Phylogenetic data analyses (IV period,
Kumpula)
n Maximum likelihood methods, Bayesian
methods, program packages
p Metabolic modelling (IV period, Kumpula)

419
Genomes sequenced – all done?
p Sequencing is just the beginning
n What do genes and proteins do?
p Functional genomics
n How do they interact with other genes
and proteins?
p Systems biology
Two sides of the same question!

420
Bioinformatics (at least mathematical
biology) can exist outside molecular biology
The metapopulation capacity of a fragmented landscape
Ilkka Hanski and Otso Ovaskainen
Nature 404, 755-758(13 April 2000)
Melitaea cinxia, Glanville Fritillary butterfly

421
Metagenomics
p Metagenomics or environmental genomics
n ”At the last count 1.8 million species were known to
science. That sounds like a lot, but in truth it's no big
deal. We may have done a reasonable job of describing
the larger stuff, but the fact remains that an average
teaspoon of water, soil or ice contains millions of micro-
organisms that have never been counted or named. ”
-- Henry Nicholls

422
Omics
p Phenome
p Exposome
p Textome
p Receptorome
p Kinome
p Neurome
p Cytome
p Predictome
p Omeome
p Reactome
p Connectome
p Genome
p Transcriptome
p Metabolome
p Metallome
p Lipidome
p Glycome
p Interactome
p Spliceome
p ORFeome
p Speechome
p Mechanome
http://en.wikipedia.org/wiki/-omics

423
Take-home messages
p Don’t trust biodatabases blindly!
n Annotation errors tend to accumulate
p Consider
n Statistical significance
n Sensitivity
of your results
p Think about the whole ”bioinformatics workflow”:
n Biological phenomenon -> Modelling -> Computation ->
Validation of results
p Results from bioinformatics tools and methods
must be validated!
p Actively seek cooperation with experts

424
Bioinformatics journals
p Bioinformatics, http://bioinformatics.oupjournals.org/
p BMC Bioinformatics,
http://www.biomedcentral.com/bmcbioinformatics
p Journal of Bioinformatics and Computational Biology
(JBCB), http://www.worldscinet.com/jbcb/jbcb.shtml
p Journal of Computational Biology,
http://www.liebertpub.com/CMB/
p IEEE/ACM Transactions on Computational Biology and
Bioinformatics , http://www.computer.org/tcbb/
p PLoS Computational Biology, www.ploscompbiol.org
p In Silico Biology, http://www.bioinfo.de/isb/
p Nature, Science (bedtime reading)

425
Bioinformatics conferences
p ISMB, Intelligent Systems for
Molecular Biology (Toronto, July 2008)
p ICSB, International Conference on
Systems Biology (Göteborg, Sweden;
22-28 August)
p RECOMB, Research in Computational
Molecular Biology
p ECCB, European Conference on
Computational Biology
p WABI, Workshop on Algorithms in
Bioinformatics
p PSB, Pacific Symposium on
Biocomputing
January 5-9, 2009
The Big Island of Hawaii

426
Master’s degree in bioinformatics?
p You can apply to MBI during the
application period November ’08 – 2
February ’09
n Bachelor’s degree in suitable field
n At least 60 ECTS credits in CS or mathstat
n English language certificate
p Passing this course gives you the first 4
credits for Bioinformatics MSc!

427
Information session on MBI
p Wednesday 19.11. 13.00-15.00 Exactum
D122
p www.cs.helsinki.fi/mbi/events/info08
p Talks in Finnish

428
Mailing list for bioinformatics courses
and events
p MBI maintains a mailing list for
announcement on bioinformatics courses
and events
p Send email to bioinfo a t cs.helsinki.fi if
you want to subscribe to the list (you can
unsubscribe in the same way)
p List is moderated

429
Computer Science
Mathematics
and Statistics
Biology & Medicine
Bioinformatics
The aim of this course
Where would you be in this triangle?
Has your position shifted during the course?

430
Feedback
p Please give feedback on the course!
n https://ilmo.cs.helsinki.fi/kurssit/servlet/Valint
a?kieli=en
p Don’t worry about your grade – you can
give feedback anonymously

431
Thank you!
p I hope you enjoyed the course!
taivasalla.net
Halichoerus grypus, Gray seal or harmaahylje in Finnish

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